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A ball is gently dropped from a height of 20m. If its velocity Increases uniformly at the rate of 10 ms^(-2) . With what velocity will it strike the ground? After what time will it strike the ground? |
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Answer» Solution :Given : height =20 m acceleration =`10 MS^(-2)` Formula : For free falling body, (i) V = gt (ii) s=1/2 `"gt"^2` time TAKEN to strike the GROUND =`s=1//2 "gt"^2` `20 m=1//2xx 10ms^(-2) xxt^2` `r^2="40 m"/(10 ms^(-2))=4s^2` `therefore` t=2s (ii)Velocity of the ball when it STRIKES ground v=gt `v=10 ms^(-2) xx2s` `v=20ms^(-1)` |
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