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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?

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?  

Grade:11

4 Answers

Piyush
askIITians Faculty 112 Points
8 years ago
Divide it into two parts
First find the value of velocity when it will going to collide with the floor let say it is v1
secondly find out the velocity INITIAL ONE for the second part taking your displacement 2.5m let say it is v2
Now 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.
Ajithesh
11 Points
7 years ago
g=9.8m/s^2
u=0
v^2/19.6=10
v=14m/s
while bouncing back
0^2-u^2/-19.6=2.5
u^2=49
u=7m/s
time in contact=0.02s
therefore, (v-u)/t=14-7/0.02=350m/s^2
a=350m/s^2
ankita patial
16 Points
5 years ago
V1=√2gh1= √2×9.8×10=√196=14
V2=√2gh2=√2×9.8×2.5=√49=7
a=F/m
F=m[v1-(v2)]/t
a=m[v1-(-v2)]/m×0.02=[14+7]/0.02=2100/2=1050
Hafsa Maheveen
13 Points
2 years ago
According to the given information
As we know
g=9.8m/s 
h1=10
h2=2.5
t=0.02 
 
a=√2gh1-(-√2gh2)/t
a=√2gh1+√2gh2/0.02
a=√2×9.8×10+√2×9.8×2.5
a=√196+√49/7
a=14+7/0.02
a=21/0.02
Therefore //a=1050//

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