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 , 9 Years ago
Grade 11
4 Answers
Piyush
Last Activity: 9 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
Last Activity: 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
Last Activity: 6 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
Last Activity: 3 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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