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`        A ball dropped on to floor from a height of 10m rebounds to a height of 2.5m. If the ball is in contact with the floor for 0.02s, its average acceleration during contact is?  `
4 years ago

Piyush
112 Points
```							Divide it into two partsFirst find the value of velocity when it will going to collide with the floor let say it is v1secondly find out the velocity INITIAL ONE for the second part taking your displacement 2.5m let say it is v2Now take t=0.02s take initial velocity to be equal to (-v1)final velocity to be equal to v2 apply the first equation of motion and find the acceleration.
```
4 years ago
Ajithesh
11 Points
```							g=9.8m/s^2u=0v^2/19.6=10v=14m/swhile bouncing back0^2-u^2/-19.6=2.5u^2=49u=7m/stime in contact=0.02stherefore, (v-u)/t=14-7/0.02=350m/s^2a=350m/s^2
```
2 years ago
ankita patial
16 Points
```							V1=√2gh1= √2×9.8×10=√196=14V2=√2gh2=√2×9.8×2.5=√49=7a=F/mF=m[v1-(v2)]/ta=m[v1-(-v2)]/m×0.02=[14+7]/0.02=2100/2=1050
```
one year 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