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A ball dropped onto the floor from a height of 10m rebounds to a height of 2.5m. If the ball is in contact with the floor of 0.02s, what is the average acceleration during contact?
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Let's take the downward velocity to be positive before striking the ground similarly let is consider velocity after striking the ground and moving upwards is to be negative
Acceleration=ratio of change in velocity to time
Velocity of a freely falling body as it hits the ground it's velocity is √2gh
If a body raises to a height h then the velocity with which it raises is also √2gh
So velocity of the ball just before hitting the ground is √2(9.8)10= 14m/s
The velocity of the ball after striking the ground is √2(9.8)2.5=7m/s
a=(v-u)/t
={14-(-7)}/0.02
=(21)/0.02
=2100/2=1050m/s2
Answer:
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