# RBSE Solutions for Class 11 Physics Chapter 3 Kinematics

## Rajasthan Board RBSE Class 11 Physics Chapter 3 Kinematics

### RBSE Class 11 Physics Chapter 3 Textbook Exercises with Solutions

#### RBSE Class 11 Physics Chapter 3 Very Short Answer Type Questions

Question 1.
What is the name of branch of Physics related with the study of motion of particles?
Dynamics.

Question 2.
How many coordinates are required there in one, two and three dimensional motion?
One dimensional motion has 1 coordinate, two dimensional motion has 2 coordinates and three dimensional motion has three coordinates.

Question 3.
In which type of motion a particle or a system of particles move around a fixed axis.
Rotational motion.

Question 4.
How much is the displacement in one cycle of a circular motion?
Zero, because initial and final points are same.

Question 5.
Write the formula of speed.
Speed = Distance/Time

Question 6.
How much would be the displacement if a person walks 4 m towards east then 3m towards the south and then 4m towards west.
3m towards south.

Question 7.
What is negative acceleration called?
Retardation.

Question 8.
What is the rate of change of displacement called with respect to time?
Velocity.

Question 9.
Name the physical quantity which obtained by the slope of velocity time graph.
Acceleration.

Question 10.
Name the physical quantity which obtained by the slope of displacement time graph.
Velocity.

Question 11.
Which physical quantity obtained by the area of velocity time graph.
Distance.

Question 12.
A particle is moving with a definite velocity then how much would be its acceleration?
Zero.

Question 13.
How much should be the angle of a projection so that the body covers the maximum distance?
45°.

#### RBSE Class 11 Physics Chapter 3 Short Answer Type Questions

Question 1.
Explain motion.
Concept of motion : If a body changes its place with time, then it is said to be in motion. An object is said to be in motion in absolute form when its speed is relative to a point in the universe which is stable. But there is no such point known in the world.

Hence, nothing is in absolute rest or in absolute motion. Truly, rest and motion both are relative. If some object relative to the other changes its position as the time passes, then the object would be said moving relative to the other object.

Question 2.
Explain in short frame of reference and its importance.
Frame of Reference
Imagine you threw and caught a ball while you were on a train moving at a constant velocity passing a station. To you, the ball appears to simply travel vertically up and then vertically down under the influence of gravity. However, to an observer standing on the station platform the ball would appear to travel in parabola, with a constant horizontal component of velocity equal to the velocity of the train. This is illustrated in figure (3.1) below. The different observations occur because the two observers are in different frames of reference.
Thus, a frame of reference is a set of coordinates that can be used to determine positions and velocities of objects in that frame. Different frames of reference move relative to one another.
Frames of reference can be of two types :
(a) Inertial frame of reference
(b) Non-inertial frame of reference.

(a) Inertial frame of reference : A frame of reference that remains at rest or moves with constant velocity with respect to other frames of reference is called inertial frame of reference. An inertial frame of reference has a constant velocity.That is, it is moving at a constant speed in a straight line, or it is standing still. Newton’s laws of motion are valid in all inertial frames of reference. Here, a body does not change due to external forces. All inertial frames of a reference are equivalent for the measurement of physical phenomena.
There are several ways to imagine this type of motion :

• Motion of Earth
• A space shuttle moving with constant velocity relative to the earth.
• A rocket moving with constant velocity relative to the earth.

(b) Non-inertial frame of reference : A frame of reference is said to be non-inertial frame of reference when a body, not acted upon by an external force, is accelerated. In a non-inertial frame of reference, Newton’s laws of motion are not valid. It also does not have a constant velocity and is accelerating. There are several ways to imagine this type of motion.

• The frame could be travelling in a straight line, but the speed of object increases or decrease.
• The frame could be travelling along a curved path at a steady speed.
• The frame could be travelling along a curved path and also the speed of object increases or decrease.
• To locate the position of a particle, the frame of reference should be universally accepted and is easily available.

The simplest reference frame is the cartesian frame of reference or cartesian coordinate system. It has three mutually perpendicular axes named as 1,7 and Z axes. The point of intersection of these axes is called origin (O) and is considered as the reference point
The X, Y and Z coordinates describe the position of the object with respect to the coordinate system. To measure time, put a clock in the system.

Question 3.
Explain and draw cartesian coordinate system.
A convenient way to fix-up the frame of reference is to choose three mutually perpendicular axis and name them x, y, z axis with a common joining point O. The coordinates x, y, z of the particle, specify the position of the particle with respect to that frame. Cartesian system is shown in the figure. Question 4.
On the basis of frame of reference in how many types is the motion divided? Write the names.
On the basis of frame of reference, the motion is divided into three types which are given below :

• One dimensional motion.
• Two dimensional motion.
• Three dimensional motion.

Question 5.
Differentiate between distance and displacement.
Distance, and displacement are two quantities that may seem to mean the same thing yet have distinctly different definitions and meanings.
(i) Distance : Distance is a scalar quantity that refers to the length of the path covered by the object regardless of its starting or ending position. In other words, distance refers to the length of the entire path travelled by the object.
Unit of distance = meter in M.K.S. system
= centimeter in C.G.S. system
Unit of displacement = meter in M.K.S. system
= centimeter in C.G.S. system.

(ii) Displacement : Displacement is a vector quantity that refers to the shortest distance between the two positions of the object i.e, the difference between the final and initial positions of the object,in a given time. Its direction is from initial to final position of the object. It is represented by the vector drawn from the initial position to its final position.

Question 6.
Explain translatory motion with examples.
Translatory motion : When an object moves from one place to other place with respect to the frame of reference, then this motion is said to be translatory motion.
Example : Motion of a car or bus on a straight road.

Question 7.
A man walks 4m east, then 5m north, then taking a right turn walks 8m straight. Calculate the distance travelled by the man and its displacement.
The distance travelled by the man
= 4 + 5 + 8 = 17 m
To determine the displacement we have to draw the vector diagram, which is in the figure. Here, AD is the displacement. The magnitude of $$\overrightarrow{A D}$$
$$|\overrightarrow{A D}|^{2}$$ = AE2 + ED2
or AD2 = (4 + 8)2 + (5)2 = (12)2 + (5)2
= 144 + 25
∴ AD = $$\sqrt{169}$$ = 13 m
or Displacement = AD = 13 m

Question 8.
Explain the differences between average and instantaneous velocity with examples.
Average velocity : Average velocity of a body is defined as the change in position or displacement (Δx) divided by time interval (Δt) in which that displacement occurs.
$$\overrightarrow{v_{a v}}=\frac{\Delta \vec{x}}{\Delta t}=\frac{\overrightarrow{x_{2}}-\overrightarrow{x_{1}}}{t_{2}-t_{1}}$$

Instantaneous velocity : The instantaneous velocity of a body is the velocity of the body at any instant of time or at any point of its path.
$$\vec { v } =\lim _{ \Delta t\rightarrow 0 } \frac { \Delta \vec { x } }{ \Delta t } =\frac { d\vec { x } }{ dt }$$
$$\vec{v}=\frac{d \vec{x}}{d t}$$
Velocity can be positive, negative or zero.
By studying speed and velocity we come to the result that at any time interval average speed of an object is equal or more than the average velocity but instantaneous speed is equal to instantaneous velocity.

Question 9.
A runner completes a circular path of 1000 m in 2 min 5 s. What is its average speed and average velocity?
Total distance travelled by the runner in 2 min 5 s = 1000 m
Total time of the whole journey t = 2 min 5 s
or t = (2 × 60) s + 5 s
= 120 s + 5 s = 125 s
Therefore average speed of the runner,
$$\vec{v}=\frac{s}{t}=\frac{1000}{125}$$ = 8m/ s v
or $$\vec{v}$$ = 8 m/s
Since, the runner runs for one complete one cycle, therefore its initial and final positions coincide with each other.
Thus, the displacement becomes zero i. e,
$$\overrightarrow{\Delta x}$$ = 0 (zero)
Hence, average velocity
$$\vec{v}=\frac{\overrightarrow{\Delta x}}{\Delta t}=\frac{0}{125}=0$$
or $$\vec{v}$$ = 0 (zero)

Question 10.
A one km long train is travelling with 2 km/min velocity. How much time will it take to cross a tunnel of 1 km in length?
Total distance travelled by the engine of the train in crossing the tunnel completely
S = 1 + 1 = 2 km
The average speed of the train
v = 2 km/min
∴ The time of crossing the tunnel,
t = $$\frac{S}{v}=\frac{2 \mathrm{km}}{2 \mathrm{km} / \mathrm{min}}$$
or t = lmin

Question 11.
Draw a velocity-time graph for a uniform accelerated motion. What does its slope depicts?
The velocity time graph for a uniform accelerated moving body is given below :
The slope of the graph represents the constant uniform acceleration. . Question 12.
A ball when thrown vertically upwards with a velocity u reaches height h. If velocity is increased 2 times (i. e. 2 u) then what would be the effect on height?
The velocity of the ball will be zero at the highest point, therefore applying the formula.
v2 = u2 + as, we have
0 = u2 – 2gh
or 2gh = u2
or h = $$\frac{u^{2}}{2 g}$$
When the initial velocity is doubled (i.e., 2v), then suppose the height becomes h’, therefore,
h’ = $$\frac{(2 u)^{2}}{2 g}=\frac{4 u^{2}}{2 g}=4 \times \frac{u^{2}}{2 g}$$
or h’ = 4h
Hence, on doubling the initial velocity, the height will become four times.

Question 13.
What is projectile motion?
Projectile motion: Projectile motion is a form of motion in which object or particle (called a projectile, is thrown near the earth’s surface; and it moves along a curved path under the action of gravity only. To study such kind of motion the velocity working on the object and the acceleration components are divided in both the horizontal and vertical directions and analyzed independently.

#### RBSE Class 11 Physics Chapter 3 Long Answer Type Questions

Question 1.
Describe the three equations for uniform accelerated motion.
Calculus method :
(i) Velocity-time relation : These equations can also be derived from calculus method. From the definition of acceleration;
a = $$\frac{d v}{d t}$$ or dv = a dt
Integrating it with in the condition of motion (i.e.) when time changes from 0 to t, velocity changes from u to v, we get This is the first equation of motion.

(ii) Distance time relation: The instantaneous velocity of an object in uniformly accelerated motion is given by
v = $$\frac{d x}{d t}$$ or dx = v dt
v = u + at
∴ dx = (u + at) dt ………….. (vii)
Let the displacement of the object from the origin of position-axis is x0 at t = 0 and x at t = t, integrating both the sides of the equation (vii) within proper limits, we have, If x – x0 = S = distance covered by the object in time t, then from eq. (viii)
s = ut + $$\frac{1}{2}$$ at2 …………… (ix)
This is the second equation of uniform accelerated motion.

(iii) Velocity-displacement relation
The instantaneous acceleration is given by Let u and v be the velocity of the object at positions given by displacements x0 and x.
Integrating the above equation (x) with the condition of motion, we get This is the third equation of uniform acceleration motion

Question 2.
Define the following :
(i) Displacement, (ii) Velocity, (iii) Acceleration, (iv) Speed, (v) Average velocity, (vi) Instantaneous velocity, (vii) Average acceleration, (viii) Instantaneous acceleration.
(i) Displacement : Displacement is a vector quantity that refers to the shortest distance between the two positions of the object i.e, the difference between the final and initial positions of the object,in a given time. Its direction is from initial to final position of the object. It is represented by the vector drawn from the initial position to its final position.

(ii) Velocity : Velocity is the vector quantity that refers to “the rate at which an object changes its position.” Imagine a person moving rapidly one step forward and one step back-always returning to the original starting position. While this might result in a frenzy of activity. It would result in a zero velocity. Because the person always returns to the original position.
The time rate of displacement is called velocity
velocity = displacement time

(iii) Acceleration: Generally, the velocity of a moving object changes with time. Sometimes the magnitude of velocity increases and sometimes it decreases.
Sometimes the magnitude remains constant but the direction changes as in circular motion. The rate of change in velocity is defined as acceleration.
Therefore,
“The rate of change of velocity of an object with respect to time is known as acceleration.”
In terms of formula: The unit of acceleration in M.K.S. system is metre/second2(m/s2) Its dimensional formula is [M0L1T-2]. Acceleration is a vector quantity. Similar to velocity it is also divided as follows:

(iv) Speed: Speed of something is the rate at which it moves or travels. Speed is defined as the rate of movement of a body expressed either as the distance travelled divided by the time taken or the rate of change of position with respect to time at a particular point. It is a scalar quantity that refers to “how fast an object is the moving.” “The time rate of distance is called speed.
Speed = $$\frac{\text { distance }}{\text { time }}$$
The unit of speed is m/s. The dimensional formula of speed is [M0LT-1]

(v) Average velocity : Average velocity of a body is defined as the change in position or displacement (Δx) divided by time interval (Δt) in which that displacement occurs.
$$\overrightarrow{v_{a v}}=\frac{\Delta \vec{x}}{\Delta t}=\frac{\overrightarrow{x_{2}}-\overrightarrow{x_{1}}}{t_{2}-t_{1}}$$

(vi) Instantaneous velocity: The instantaneous velocity of a body is the velocity of the body at any instant of time or at any point of its path.
$$\vec{v}=\lim _{\Delta t \rightarrow 0} \frac{\Delta \vec{x}}{\Delta t}=\frac{d \vec{x}}{d t}$$
$$\vec{v}=\frac{d \vec{x}}{d t}$$
Velocity can be positive, negative or zero.
By studying speed and velocity we come to the result that at any time interval average speed of an object is equal or more than the average velocity but instantaneous speed is equal to instantaneous velocity.

(vii) Average Acceleration: “The ratio of total change in velocity to the total time taken is called average acceleration”.
If Δv is the change in velocity in Δt time interval, then;
Average acceleration (viii) Instantaneous Acceleration : Instantaneous acceleration is defined as “acceleration at any given point or at any instant of time.” If at Δt time interval velocity is Δv then according to above definition; to calculate instantaneous acceleration Δt → 0. Hence,
Instantaneous Acceleration $$a=\lim _{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}=\frac{d v}{d t}$$
Here, $$\frac{d v}{d t}$$, differentiation of v with respect to time dt t which can be known mathematically.
∵ v = $$\frac{d x}{d t}$$
Therefore, $$a=\frac{d}{d t}\left[\frac{d x}{d t}\right]=\frac{d^{2} x}{d t^{2}}$$
Here, $$\frac{d^{2} x}{d t^{2}}$$, double differentiation of x w.r.t. t
which can he calculated mathematically.
Therefore, instantaneous acceleration is differentiation of velocity with respect to time time and is double differentiation of displacement w.r.t. time.

For any moving object at definite time intervals, if the change in velocity is also same then this is known as uniform acceleration. And if the changes are different in velocity then this is non-uniform acceleration. In same accelerated motion average acceleration and instantaneous acceleration are same.

If in any circular motion (path), the magnitude of velocity of the moving object does not change but the direction of the moving object changes continously, they this type of motion is also called accelerated motion.

Acceleration can be positive, negative or zero. If acceleration is positive its velocity increases. If acceleration is zero then the object moves with a constant speed (velocity). And if the acceleration is negative then the velocity of the object decreases. Hence, negative acceleration is called retardation.

Question 3.
Derive the equations of motion with the help of the graphical method.
Graphical method : Consider an object moving along a straight line with uniform acceleration a. Let u be the initial velocity of the object at time t = 0 and v be the final velocity of the object at time t. Let s be the distance travelled by the object in time t. Velocity-time graph of this motion is a straight line PQ, as shown in the figure (3.13).
where, OP = u = RS
OW = SQ = v and
OS = PR = t

(i) For the first equation of motion: We know that the slope of velocity-time graph of uniformly accelerated motion represents the acceleration of the object.
i.e., Acceleration = slope of the velocity-time graph PQ This is the first equation of uniform acceterated motion.

(ii) Second equation of motion : We know that the area under the velocity-time graph for a given time interval represents the distance covered by the uniformly accelerated object in that interval of time.
∴ Distance (displacement) travelled by the object in time t is :
S = area of trapezium OSQP
= area of rectangle OSRP + Area of triangle PRQ
or S = OS × OP + $$\frac{1}{2}$$ × PR × RQ
(Area of rectangle = Length × Breadth)
(Area of triangle = $$\frac{1}{2}$$ × Base × Height)
= t × u + $$\frac{1}{2}$$ × t × (v – u)
(From the first equation of motion v – u = at)
= ut + $$\frac{1}{2}$$ × t × at
Thus, S = ut + $$\frac{1}{2}$$ at2 ………….. (ii)
This is the second equation of uniform accelerated motion

(iii) Third equation of motion : Distance travelled by the object in time interval t is
s = area of trapezium OSQP
= $$\frac{1}{2}$$ (OP + SQ) × OS
∵ OP = SR
= $$\frac{1}{2}$$ (SR + SQ) × OS ……………. (iii)
Acceleration, a = slope of the velocity-time graph PQ This is the third equation of uniform accelerated motion.

Question 4.
Prove that the path of projectile motion is parabolic.
Path of a projectile
Let OX be a horizontal line on the ground and OY be a vertical line: O is the origin for X and Y axis.

Consider that a projectile is fired with velocity u and making an angle θ with the horizontal from the point ‘O’ on the ground [figure 3.17], ’ The velocity of projection of the projectile can be resolved into the following two components
(i) ux = u cosθ, along OX
(ii) uy = u sinθ, along OY .
As the- projectile moves, it covers distance along the horizontal due to the horizontal component u cosθ of the velocity of projection and along vertical due to the vertical component u sinθ. Let that any time t, the projectile reaches the point P, so that its distances along the X and Y-axis are given by x and y respectively.

Motion along horizontal direction : It we neglect the friction due to air, then horizontal component of the velocity i. e., u cosθ will remain constant. Thus
Initial velocity along the horizontal, ux = u cosθ
Acceleration along the horizontal, ax = 0
The position of the projectile along X-axis at any time t is given by
x = uxt + $$\frac{1}{2}$$ axt2
Putting ux= u cosθ and ax = 0, we have
x = (u cosθ)t + $$\frac{1}{2}$$ (0)t2
or x = (ucosθ)t
or t = $$\frac{x}{u \cos \theta}$$ ……………… (i)

Motion along vertical direction :
The velocity of the projectile along the vertical goes on decreasing due to effect of gravity
Initial velocity along vertical, uy = u sinθ
Acceleration along vertical, ay = -g
The position of the projectile along T-axis at any time t is given by This is an equation of a parabola. Hence the path of a projectile projected at some angle with the horizontal direction is a parabola.

Question 5.
Derive formulae for time of flight (T), maximum height (H) and horizontal range (R) of projectile motion.
Time of Flight
It is the total time taken by the projectile when it is projected from a point and reaches the same horizontal plane or the time for which the projectile remains in the air above the horizontal plane.
It is denoted by T.

As the motion from the point O to A and then from the point A to B are symmetrical, the time of ascent (for journey from point O to A) and the time of descent (for the return journey from A to B) will be each equal to T/2.

At the highest point A, the vertical component of velocity of the object becomes zero. Taking vertically upward motion of the object from O to A,we have
uy = u sinθ, ay = -g, t = $$\frac{T}{2}$$ and vy = 0
since, vy = uy + at
∴ 0 = u sinθ – g $$\frac{T}{2}$$
⇒ T = $$\frac{2 u \sin \theta}{g}$$ …………….. (4)

Maximum height of a projectile
It is the maximum vertical height attained by the object above the point of projection during its flight. It is denoted by H.

Taking the vertical upward motion of the object 122 (B )from O to A, we have : Horizontal-range
It is the horizontal distance covered by the object between its point of projection and the point of hitting the ground. It is denoted by R.

Obviously, the horizontal range R is the horizontal distance covered by the projectile with the’ uniform velocity u cosθ in a time equal to the time of flight. Question 6.
Derive formula for displacement, velocity and acceleration of two-dimensional motion.
Displacement, Velocity, and Acceleration of a Particle in Two Dimensional Motion and Their Vector Representation
To study about displacement, velocity and acceleration of a particle in two dimensional motion we study the about the motion of the particle in XY axis of the reference frame. Assume that at any time interval t1 the position of the particle is A1 whose position vector is $$\overrightarrow{r_{1}}$$ and at any time t2 the position of the particle is A2 whose position vector is $$\overrightarrow{r_{2}}$$. Therefore, the vector representation of points A1 and A2 is following; Where $$\hat{i}$$ and $$\hat{j}$$ are unit vectors in the direction of x and y axis representing directions respectively.
and $$\vec{r}_{2}=x_{2} \hat{i}+y_{2} \hat{j}$$ …………… (2)
Again, because the particle is displaced from A1 to A2. Hence, according to the diagram if displacement is Δr then according to the triangle rule in the triangle OA1A2. Therefore, the displacement equation would be given as equation (3) in two dimensional motion.
According to Vector Algebra the resultant is;
Δr = $$\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}$$ ………………… (4)
This equation (3) is the result of displacement and the direction of displacement vector Δr would be according to the figure (3.15).

To calculate the average velocity of the particle according to the definition of average velocity; Hence, average velocity would be given as equation (5) and the resultant of average velocity by
Vector Algebra would be;
v = $$\sqrt{\left(v_{x}\right)^{2}+\left(v_{y}\right)^{2}}$$ ……………. (6)
given by equation (6) and the direction of it would be in the direction of $$\overrightarrow{\Delta r}$$. The instantaneous velocity of this particle at any instant t by the definition of instantaneous velocity would be; Hence, instantaneous velocity is given by the eqn. (7). Resultant of instantaneous velocity by Vector Algebra.
v = $$\sqrt{v_{x}^{2}+v_{y}^{2}}$$ …………….. (8)
will be given by eqn. (3) and direction would be in the direction of the tangent at the point; which is the position of the particle at time t. Suppose it makes an angle θ with the x-axis; then the direction would be as shown in the figure (9) and by the equation;
tan θ = $$\frac{v_{y}}{v_{x}}$$
or θ = $$\tan ^{-1}\left[\frac{v_{y}}{v_{x}}\right]$$ ……………….. (9)
If the object is executing accelerated motion and in the time interval ∆t the change in velocity is ∆v, then by the definition of average acceleration, the acceleration of the particle will be; This is the equation of average acceleration.
Result of this by Vector Algebra is given by
a = $$\sqrt{\left(a_{x}\right)^{2}+\left(a_{y}\right)^{2}}$$ ………………. (11)
will be given by equation (11) and the direction of it would be in the direction of ∆v.
To calculate instantaneous acceleration by Equation (12) is the vector equation of instantaneous acceleration; and its result;
a = $$\sqrt{\left(a_{x}\right)^{2}+\left(a_{y}\right)^{2}}$$ …………… (13)
will be given by (13) and its direction would be in the direction of v. All the various equations above explain displacement, velocity and acceleration in two-dimensional motion.

#### RBSE Class 11 Physics Chapter 3 Numerical Problems

Question 1.
Time taken from Jaipur to Ajmer by car is 1.5 h. If the average velocity of car is 80 km/h then how much is the distance between Jaipur and Ajmer?
Solution:
Given; time t = 1.5 hr.; $$\overline{v}$$ = 80km/hr
s = ?
∵ $$\overline{v}=\frac{s}{t} \Rightarrow s=\overline{v} \times t$$
∴ s = 80 × 1.5 = 120 km

Question 2.
In 5s the speed of bus increases from 25km/h to 70km/h. What is the average acceleration of the bus?
Solution:
∵ Average acceleration Question 3.
Car A and B travels 100 km journey together. Car A travels with the uniform speed of 40 km/h. Car B travels with 60 km/h. But after half an hour it stops for 15 min due to some problem and when it moves its speed remains 50 km/h.
(i) Draw a speed-time graph of the journey.
(ii) Tell which car will complete the journey first and how much time before?
Solution:
Given; l = 100km; vA = 40 km/hr; vB = 60 km/hr; v’B = 50 km/hr.
For complete journey, the time taken by car A,  Thus tA> tB
Obviously, car B will complete its journey first. The difference of time between two cars,
Δt = tA – tB = (2.5 – 2.15)hr.
= 0.35 hr.
= 0.35 × 60 min = 21 min
Thus car B will reach the destination 21 min. before car A
The speed time graph for the cars will be shown as below.

Question 4.
The displacement of a moving particle in effect of constant force is given as :
t = $$\sqrt{x}+3$$, where x is displacement (in m) and t is time (in s). When its velocity is zero, calculate the displacement.
Solution: Therefore the displacement of the particle will zero when its velocity is zero.

Question 5.
A particle is dropped from a 200 m tall tower. At the same time a second particle is thrown upwards with velocity 50 m/s. Calculate when and where would these two particles (bodies) meet.
Solution:
Suppose both the bodies y meet at point C. Now, the distance h travelled by the body A which is dropped freely from the top of the tower, using relation,  ∴ The height of meeting point of both the bodies from the ground.
= 200 – h = 200 – 78.4 = 121.6 m
Thus both the bodies will meet after 4s at the height of 121.6 m from the ground.

Question 6.
A 100 g mass particle falls 490 cm downwards from rest position and comes to rest after moving 70 cm down in the sand. Calculate
the acceleration imposed by the sand on the particle. (g = 9.8m/s2)
Solution:
Given; = 100g = 0.1kg
h = 490cm = 4.90 m
The distance of penetration in sand s = 70 m = 0.70 m
Acceleration imposed by sand on the particle,
a =?
Now applying the formula;
v2 = u2 + 2as
We get
v2 = 0 + 2gh = 2 × 9.8 × 4.9
or v2 = 2 × 2 × 4.9 × 4.9
∴ v = 2 × 4.9 = 9.8 ms-1
When the particle enters the sand and comes in rest, then
u = 9.8 m/s; v = 0; a = ?; s = 0.70 m.
Applying the formula v2 = u2 + 2as, we have
0 = (9.8)2 – 2a × 0.7
or 2 × a × 0.7 = 9.8 × 9.8
∴ a = $$\frac{9.8 \times 9.8}{2 \times 0.7}$$ = 4.9 × 14
or a = 68.6 ms-1

Question 7.
A stone is dropped from a helicopter moving in upward direction at 500 m height. It reaches the earth after 6 seconds. What was the velocity of the helicopter when the stone was dropped? (g = 10m/s2)
Solution:
Given; h = 500m; t = 6s; u = ?; g = 10 m/s
On using relation s = ut + $$\frac{1}{2}$$ at2 we have, Therefore, the velocity stone = 53.33 m/s in, upward direction.

Question 8.
A man walks with a speed of 5 km/h on a straight road and reaches his office which is 2.5 km away. Since, the office was closed, he at the same time returns back to his house with a speed of 7.5 km/h. Calculate his average speed and average velocity for following time intervals.
(i) 0 to 30 min
(ii) 0 to 50 min
(iii) 0 to 40 min. Solution:
(i) The time taken from home to office, (ii) For the interval 0 – 50 min
In returning journey, upto 50 min he will walk for (50 – 30) = 20 min.
Therefore distance travelled in this interval of time,
s’ = v’ × t’ = 7.5 × $$\frac{20}{60}$$ km
s’ = 2.5km
Thus, in next 20 min, he will reach his home.
∴ Travelled distance = 2.5 + 2.5 =5 km
Total time 50 min = 50/ 60 = 5/ 6hr
Total displacement = 0
Hence, the average velocity = $$\frac{0}{5 / 6}$$ = 0 km/hr
and average speed = $$\frac{\text { Totaldistance }}{\text { Total time }}=\frac{5}{5 / 6}$$
= 6 km/hr

(iii) For time interval 0 to 40 min
Time of return journey = 40 – 30 = 10 min
= $$\frac{10}{60} \mathrm{hr}=\frac{1}{6} \mathrm{hr}$$
The distance travelled in this time = 7.5 × 1/ 6 = 1.25 km
Total time = 40 min = $$\frac{40}{60}=\frac{2}{3}$$ km
Total displacement = 2.5 – 1.25 = 1.25 km.
Hence, average speed. Question 9.
A man walks 18 m east in 16 s and turns north. Now walks 5 m north in 5 s and then turns left and walks 6 m in 8 s and stops. Calculate the average speed and average velocity during the journey of the man.
Solution:
Complete journey of the man is shown in following vector the figure. Total time of the journey,  Question 10.
A uniform accelerated motion object covers 65 m in 5th second and 105 m in 9th. Then how much distance will it cover in 20th second? Also calculate the distance travelled in 20 s.
Solution:
The distance travelled in nth second, Subtracting equation (1) from equation (2), we get Question 11.
A bullet is fired from a gun at an angle of 30° from the horizontal; which falls 3 km away on the ground. Assuming the air resistance negligible and the gun’s tunnel velocity unchanged; can it be used to fire a 5 km distant object?
Solution:
Given; θ = 30° ; R = 3km = 3000 m; ∵ $$\frac{u^{2}}{g}$$ is unchanged and the maximum value of sin 2θ is 1. Therefore, the gun cannot be use for the range of 5 km.

Question 12.
Calculate the projection angle of a projectile whose range is 50 m and maximum height 10m.
Solution:
Given; R = 50m; A = 10m; θ = ? Question 13.
Prove that the maximum horizontal range is 4 times of the height attained by the particle.
Solution:
∵ Range R = $$\frac{u^{2} \sin 2 \theta}{g}$$
and maximum height H = $$\frac{u^{2} \sin ^{2} \theta}{2 g}$$ or R = H $$4 \frac{\cos \theta}{\sin \theta}$$ = 4H. cotθ
∵ For maximum range θ = 45°
∴ cot θ = 1
∴ R = 4H