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The height from which a body is released numerically equal to the velocity aquired finally then that height is equal to

The height from which a body is released numerically equal to the velocity aquired finally then that height is equal to

Grade:12

1 Answers

Rohan
23 Points
6 years ago
As s= ut + a(t)(t)/2 but in this case u=0 as the body is dropped simply.Also v=u +at.Putting u=0 in both the cases we get, s=a(t)(t)/2 and v=at.As given s=v ,and cancelling a&t from both sides we get t=2 sec.Now a=g ,t=2 hence s=9.8× 4/2=19.6m.Hence height= 19.6 m.

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