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Grade: 11
        
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

Answers : (3)

Piyush
askIITians Faculty
112 Points
							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.
4 years ago
Ajithesh
11 Points
							
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
2 years ago
ankita patial
16 Points
							
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
one year ago
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