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From the top of a tower of height 40m, a ball is projected upward with a speed of `20ms^(-1)` at an angle of elevation of `30^(@)`. Then the ratio of the total time taken by the ball to hit the ground to the time taken to ball come at same level as top of tower.A. `2:1`B. `3:1`C. `3:2`D. `4:1` |
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Answer» Correct Answer - a If t is the total time taken, then `40=-20sin 30^(@)t+1/2xx10xxt^(2)` or `40=-10t+5t^(2)` or `5t^(2)-10t-40=0` or `t^(2)-2t-8=0` or `t^(2)-4t+2t-8=0` or `t(t-4)+2(t-4)=0` or `(t+2)(t-4)=0` t=4s [Negative time is not allowed] `T=(2v sin theta)/g=(2xx20sin 30^(@))/10s=2s` `:. t/T=4/2=2/1` |
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