__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/s

^{2}

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)at

^{2}

90 = 0 + (1/2) X 9.8 t

^{2}

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,

*u*

_{r}= 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*=

*u*

_{r}+

*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:

## No comments:

## Post a Comment