1.

A ball is released from the top of a tower of height `H m`. After `2 s` is stopped and then instantaneously released. What will be its heitht after next `2 s`?.A. `(H-5) m`B. ` (H-10) m`C. ` (H-20) m`D. `(H-40) m`

Answer» Correct Answer - D
Distance covered by by the object in first `2 s`
`h_(1)=(1)/(2) g t^(2) =(1)/(2) xx10xx2^(2) =20 m`
Similarly , destance covered by the object invext `2 s` will also be `20 m`, hence the required height `=H-20-20=H-40 m`.


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