NCERT solution Class 11 motion in a straight line

NCERT Solutions

Class 11 Physics

Motion In A Straight Line

NCERT Questions

Q. 2.1
Q. 2.2
Q. 2.3
Q. 2.4
Q. 2.5
Q. 2.9
Q. 2.6
Q. 2.7
Q. 2.8
Q. 2.10
Q. 2.11
Q. 2.12
Q. 2.13
Q. 2.14
Q. 2.15
Q. 2.16
Q. 2.17
Q. 2.18

Q. 2.8

A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.

Given:-

Ball is dropped from a height, s =90m

Initial velocity of the ball, u = 0

Acceleration, a = g = 9.8 m/s

Final velocity of the ball = v

From second equation of motion,

s = ut+ (1/2) at²

90=0 + (1/2) x 9.8(t²)

t = √18.38 = 4.29s

From first equation of motion, final velocity is given

v = u + at

= 0 + (9.8 × 4.29) = 42.04 m/s

Rebound velocity of the ball, u_r = (9/10) v= (9/10) x 42.04 =37.84m/s

Time (t) taken by the ball to reach maximum height can be obtained by using First equation of motion as:

v = u_r+ at′

0 = 37.84 + (– 9.8) t′

t’ = (-37.84)/(-9.8) = 3.86s

Total time taken by the ball = (t + t’) = (4.29+3.86) =8.15s.

As the time of ascent is equal to the time of descent, the ball takes 3.86 s to strike back on the floor for the second time.

The velocity with which the ball rebounds from the floor = ((9/10) x 37.84) =34.05m/s

Total time taken by the ball for second rebound = 8.15 + 3.86 = 12.01 s

The speed-time graph of the ball is represented in the given figure as:

Video solution:

Motion In A Straight Line Class 11 Physics NCERT Chapter 3 NCERT Solutions

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NCERT solutions of Motion In A Straight Line Class 11 Physics - all questions of all exercises are solved. Q. No. 1, 2, 3, and so on. Complete explanation and detailed solution makes them the best videos for Class 11 Physics Motion In A Straight Line NCERT solutions.