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An object is dropped from the top of a building 5 metres and rebounds to a height of 3.2 metre if it is in contact with the floor for 0.036 second, what is its average acceleration during this period ? (g=10m per sec sq)

```
2 years ago

Arun
25768 Points
```							While falling down, initial velocity = 0, final velocity = v, acceleration due to gravity = 9.8 m/s2, height = 5 mThe velocity of the ball with which it hits the ground can be found using,v2 = u2 + 2as=> v2 = 0 + 2×9.8×5=> v = 9.9 m/s (downward)While it moves up, initial velocity = u, final velocity = 0, acceleration due to gravity = 9.8 m/s2, height = 3.2 mThe velocity of the ball with which it leaves the ground can be found using,v2 = u2 + 2as=> 0 = u2 - 2×9.8×3.2 [this time the velocity and acceleration are oppositely directed so we have the negative sign]=> v = 7.91 m/s (upward)So, Change in velocity = 7.91 m/s (upward) – 9.9 m/s (downward)=> Δv = 7.91 m/s + 9.9 m/s  [both in upward direction]=> Δv = 17.81 m/sSo, acceleration is, a = Δv/Δt = 17.81/0.036 = 494.7 m/s2
```
2 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions