Question

If a ball of mass 0.1 kg hits the ground from the height of 20m and bounce back to the same height, then find out the force exerted on the ball if the time of impact is 0.04(Takc, g = 10 m/s ^ 2)​

Answer 1

Answer:

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Explanation:

Mass of the ball m = 0.1 kg

Time in contact with the ground, t = 0.01s

1. When the ball is dropped on ground,

u =0, s = 2.5m, g=10 ms

−2

From the equation, v

2

=u

2

+2gs

v

2

=0+2×10×2.5=7.07 ms

−1

, downwards.

2. When the ball rises up from the ground,

v = 0, s = 0.4m, g=10 ms

−2

From the equation v

2

=u

2

+2gs

0=u

2

+2(−10)0.4

u

2

=8=2.83 ms

−1

, upwards.

Assuming the upward velocity +ve & downward velocity -ve,

change of velocity = 2.83 - (- 7.07)

i.e v - u = 9.9 ms

−1

∴ Force exerted by the ground on the ball is,

F=m(

t

v−u

)=0.1(

0.01

9.9

)=99N.