Four bodies P, Q, R and S are projected with | vedclass

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(A) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the initial velocity,$	heta$ is t

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$P$

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(A) The horizontal range $R$ of a projectile is given by the formula $R = \frac{u^2 \sin(2	heta)}{g}$,where $u$ is the initial velocity,$	heta$ is the angle of projection,and $g$ is the acceleration due to gravity.Since all bodies are projected with the same velocity $u$,the range depends on $\sin(2	heta)$.For $P$: $	heta = 15^{\circ}$,so $2	heta = 30^{\circ}$,$\sin(30^{\circ}) = 0.5$.For $Q$: $	heta = 30^{\circ}$,so $2	heta = 60^{\circ}$,$\sin(60^{\circ}) \approx 0.866$.For $R$: $	heta = 45^{\circ}$,so $2	heta = 90^{\circ}$,$\sin(90^{\circ}) = 1$.For $S$: $	heta = 60^{\circ}$,so $2	heta = 120^{\circ}$,$\sin(120^{\circ}) = \sin(60^{\circ}) \approx 0.866$.Comparing the values,$\sin(30^{\circ})$ is the smallest. Therefore,body $P$ has the shortest range.

Four bodies $P, Q, R$ and $S$ are projected with equal velocities having angles of projection $15^{\circ}, 30^{\circ}, 45^{\circ}$ and $60^{\circ}$ with the horizontal respectively. The body having the shortest range is

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