1.

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?

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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