Asked by patel
a ball is dropped from height 10 m. it rebounds to a height of 5m.the ball was in contact with the floor for 0.01s what was its acceleration during contact?
Answers
Answered by
Abhinav Verma
u = root ( 2gh)
u = root(2 * 9.8 * 10)
u = root(196
u = 14 m / s
v = root(2gh)
v = root(2*9.8 * 5)
v = root(98)
v = root 98
v = 7 root2
14 + 7 root2
Acceleration = (14+7root2)/ 0.01
= 7(2 + root2) / 0.01
= 700(2 + root2)
700(2 + 1.4142)
= 700(3.4142)
= 341.42 *7
= 2389.94 m/sec^2
u = root(2 * 9.8 * 10)
u = root(196
u = 14 m / s
v = root(2gh)
v = root(2*9.8 * 5)
v = root(98)
v = root 98
v = 7 root2
14 + 7 root2
Acceleration = (14+7root2)/ 0.01
= 7(2 + root2) / 0.01
= 700(2 + root2)
700(2 + 1.4142)
= 700(3.4142)
= 341.42 *7
= 2389.94 m/sec^2
Answered by
Anonymous
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