Question
A particle is projected from a point $O$ with a velocity $u$ in a direction making an angle $\alpha$ upward with the horizontal. After some time at point $P$ it is moving at right angle with its initial direction of projection. The time of flight from $O$ to $P$ is (a) $\frac{u \sin \alpha}{g}$(b) $\frac{u \operatorname{cosec} \alpha}{g}$(c) $\frac{u \tan \alpha}{g}$(d) $\frac{u \sec \alpha}{g}$
Step 1
Therefore, we can resolve this velocity into two components: horizontal and vertical. The horizontal component of the velocity is $u \cos \alpha$ and the vertical component is $u \sin \alpha$. Show more…
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Motions in Two and Three Dimensions
Section B
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