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A ball is dropped on the floor from a height 10 m . It rebounds to a height of 2.5 m .If the ball is contact with floor for 0.01 sec , then average acceleration during contact is?
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A ball is dropped on the floor from a height 10 m . It rebounds to a h...
If a ball falls through height "h" (or) rises to a height "h" it's velocity is given by √2gh. So,
velocity before collision = √2(10)(10) = 10√2
velocity after collision = √2(10)(2.5) = 5√2
change in momentum = impulse (J)
On fixing the upward direction positive we get impulse as,
J = m(5√2) - m(-10√2) = 15m√2
Now, J = F∆t
⇒15m√2 = (ma)(0.01)
⇒a = 15√2 × 10² m/s²
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A ball is dropped on the floor from a height 10 m . It rebounds to a height of 2.5 m .If the ball is contact with floor for 0.01 sec , then average acceleration during contact is?
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