InterviewSolution
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InterviewSolution
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A ball is thrown from the top of tower with an initial velocity of `10ms^(-1)` at an angle of `30^(@)` with the horizontal. If it hits the ground of a distance of 17.3m from the back of the tower, the height of the tower is `(Take g=10ms^(-2))`A. `5m`B. `20m`C. `15m`D. `10m` |
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Answer» Correct Answer - d Here, `theta=30^(@), u=10ms^(-1)`, `R=17.3m, g=10ms^(-2)` For horizontal motion, `R=u cos thetat` `t=R/(u cos theta)=17.3/(10 cos 30^(@))=(17.3xx2)/(10xxsqrt(3))=(17.3xx2)/(10xx1.73)=2s` For vertical motions, `h=u sin thetat=1/2"gt"^(2)` `=10sin 30^(@)xx2-1/2xx10xx2^(2)` `=10-20=-10m`. Height of tower =10m. |
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