InterviewSolution
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InterviewSolution
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Particle projected from tower fo heigh 10m as shown in figure. Find the time (in sec) after which particle will hit the ground. |
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Answer» Correct Answer - 2 `u_(y)=v sin 45^(@)=5sqrt(2)xx(1)/sqrt(2)=5ms^(-1)` Using, `y=u_(y)t+(1)/(2)a_(y)t^(2)` `-10=5t+(1)/(2)(-10)t^(2)` `t^(2)-t=2=0` `(r-2)(t+1)=0` `therefore` t=2 (rejecting -ve time) |
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