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Grade 11Mechanics

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?

Raghu Vamshi Hemadri

11 Years agoGrade 11

4 Answers

Piyush

11 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

9 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

7 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

5 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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