Q. 123.9( 78 Votes )

# 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.

Answer :

Given,

Ball is dropped from a height = 90 m

Time interval, 0 ≤ t ≤ 12

Initial velocity of the ball =0 m/s

Final velocity of the ball = v m/s

Acceleration due to gravity, g = 9.81 ms^{-2}

From 2^{nd} equation of motion for freely falling body,

s = ut + 0.5gt^{2}

where,

u = Initial velocity

g = Acceleration due to gravity

s = Distance covered

t = Time

⇒ 90 = 0 + (0.5×9.81)t^{2}

∴ t = 4.29 s

From 1^{st} equation of motion for freely falling body,

v = u + gt

where,

v = Final velocity

u = Initial velocity

g = Acceleration due to gravity

t = Time

⇒ v = 0 + (9.81×4.29) = 42.04 m/s

Bounce velocity of the ball, v_{b} = 0.9v = 37.84 m/s

Time (t’) by the bouncing ball to reach maximum is given by,

v = v_{b} – gt’

where,

v = Final velocity

v_{b} = Bounce velocity

g = Acceleration due to gravity

t’ = Bouncing time

⇒ 0 = 37.84 – (9.81×t’)

∴ t’ = 3.86 s

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

The iterations go on like this up to ball reach a static condition.

The graph obtained by the data is,

Rate this question :

A ball falls on the ground from a height of 2.0 m and rebounds up to a height of 1.5 m. Find the coefficient of restitution.

HC Verma - Concepts of Physics Part 1