A ball is dropped on the floor from a height of 10 M it rebounds to a height of 2.5 M if the ball is in contact with the floor for 0.01 seconds then average acceleration due during contract is?
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Here,
Height from which the ball is dropped is, h = 10 m
Velocity with which the ball hits the ground can be found as,
v2 = u2 + 2gh
=> v2 = 2gh
=> v = (2gh)1/2 [downward]
The ball then rebounds to a height of, h/ = 2.5 m. Let the velocity with which the ball rebounds be v/.
So, using,
02 = (v/)2 – 2gh/
=> v/ = (2gh/)1/2 [upward]
The time for which the ball was in touch with the ground is t = 0.01 s
So, acceleration of the ball is,
a = (v/ - v)/t
[considering upward velocity to be positive, v/is positive and v is negative]
=> a = [(2gh/)1/2 – {-(2gh)1/2}]/t
=> a = [(2×9.8×2.5)1/2 + (2×9.8×10)1/2]/0.01
=> a = 2100 m/s2
Height from which the ball is dropped is, h = 10 m
Velocity with which the ball hits the ground can be found as,
v2 = u2 + 2gh
=> v2 = 2gh
=> v = (2gh)1/2 [downward]
The ball then rebounds to a height of, h/ = 2.5 m. Let the velocity with which the ball rebounds be v/.
So, using,
02 = (v/)2 – 2gh/
=> v/ = (2gh/)1/2 [upward]
The time for which the ball was in touch with the ground is t = 0.01 s
So, acceleration of the ball is,
a = (v/ - v)/t
[considering upward velocity to be positive, v/is positive and v is negative]
=> a = [(2gh/)1/2 – {-(2gh)1/2}]/t
=> a = [(2×9.8×2.5)1/2 + (2×9.8×10)1/2]/0.01
=> a = 2100 m/s2
hi mareeva
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