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A tennis ball is dropped on to the floor from a height of 4m.it rebounds to a height of 3m.if the ball was in contact with floor for 0.010 sec, what was its average acceleration during contact. acceleration=[(2gh)^1/2]+[(2gh)^1/2]/t=[(2*9.8*3)^1/2]+[(2*9.8*4)^1/2]/0.01=1652m/s

i have doubt in while using time.in the expression for acceleration "t" is the time required to travel the distance.so in that case acceleration will be=[velocity of fall+velocity of rebound/time required for falling+rebounding]

but here we used "time in ball in contact with the floor" instead of total time. what is concept/logic behind it.

Please guide me the correct way to achieve my objective.

IntelliJ Amiya
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user2445382
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1 Answers1

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For the falling ball V^2 = 2*g*h so V at first contact with ground is sqrt(78.4) = 8.85 m/s downwards. For the rising ball, the same argument can be used to show V = 7.67 m/s upwards. So deltaV is 8.85+7.67 = 16.52 m/s.

But deltaV = a*t so a = deltaV/t = 16.52/0.01 = 1652 m/s (equal to ~168.6 times gravitational acceleration).

I would suggest that if you want to understand this a little better, you should begin by searching for a tutorial explaining kinematics.

Penguino
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