Asked by Anonymous

After falling from rest from a height of 27 m, a 0.53 kg ball rebounds upward, reaching a height of 17 m. If the contact between ball and ground lasted 1.6 ms, what average force was exerted on the ball?
Answered by AI
Vf = sqrt (2 * 9.8* 27m) = 23 m/s

Vi = sqrt (2 * 9.8* 17m) = 18.25 m/s

a = change in velocity/ change in time
= Vf - Vi/ tf - ti

= [23 - (-18.25)] / [.0016 s - 0s]

= 41.25 / .0016 = 25781.25 m/s^2

you know F = ma

so F = .53kg * 25781.25 m/s^2

F = 13664 N

The reason why the velocity is negative in after the bounce I think is because it is now moving downward.

The answer is correct though for sure.
Had the same problem to work on :p

Answered by LOL
ACTUALLY IT IS BECAUSE THE FINAL VELOCITY BECOMES THE INITIAL VELOCITY IN THE SECOND ECUATION AND VICEVERSA, SO IT SHOULD BE [18.25 - (-23)] / [.0016 s - 0s] , BECAUSE THE NEGATIVE V IS 23...
Answered by Emily
This is wrong!
Answered by GG
This was correct for me, thank you!
Answered by James
How did you guys find time?
Answered by L
It gives you time, convert 1.6 ms to seconds.
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