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### Motion in a straight line NCERT Solutions Class 11 Physics - Solved Exercise Question 3.12

Question 3.12

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.

Solution:

Ball is dropped from a height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = 9.8 m/s2
Final velocity of the ball = v
From second equation of motion, time (t) taken by the ball to hit the ground can be obtained as:
s = ut + (1/2)at2
90 = 0 + (1/2) X 9.8 t2

t = √18.38  = 4.29 s
From first equation of motion, final velocity is given as:
v = u + at
= 0 + 9.8 × 4.29 = 42.04 m/s
Rebound velocity of the ball, ur = 9v / 10 = 9 X 42.04 / 10 = 37.84 m/s
Time (t) taken by the ball to reach maximum height is obtained with the help of first equation of motion as:
v = ur + at
0 = 37.84 + (– 9.8) t
t′ = -37.84 / -9.8 = 3.86 s
Total time taken by the ball = t + t′ = 4.29 + 3.86 = 8.15 s
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 X 37.84 / 10 = 34.05 m/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: