Question
Asked Nov 2, 2019

After falling from rest from a height of 25 m, a 0.30-kg ball rebounds upward, reaching a height of 15 m. If the contact between ball and ground lasted 2.1 ms, what average force was exerted on the ball? (Enter the magnitude.) 

check_circleExpert Solution
Step 1

Given information:

The mass of the ball (m) = 0.3 kg

Height from which it falls (h1) = 25 m

Height to which it rebounds (h2) = 15 m

The time for which the force acts (Δt) = 2.1 ms

Consider the down ward direction as “-ve” and upward direction as “+ve”

Step 2

The velocity (u) at which it reaches the ground wh...

u2 = 2gh1
2(9.81) (25) = 22.14 m/s
=
u 22.14 m/s downwards (or) -22.14 m/s
The velocity (v) at which it re-bounces from the ground and reaches to height h2 is given by:
/2gh2 = 2(9.81) (15) 17.15 m/s
v2
17.15 m/s upwards (or) 17.15 m/s
v
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u2 = 2gh1 2(9.81) (25) = 22.14 m/s = u 22.14 m/s downwards (or) -22.14 m/s The velocity (v) at which it re-bounces from the ground and reaches to height h2 is given by: /2gh2 = 2(9.81) (15) 17.15 m/s v2 17.15 m/s upwards (or) 17.15 m/s v

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