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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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