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 $u$ and $g$ are constant for all bodies,the range $R$ depends on $\sin(2	heta)$.For the given angles:For $P$: $	heta = 15^o$,$2	heta = 30^o$,$\sin(30^o) = 0.5$For $Q$: $	heta = 30^o$,$2	heta = 60^o$,$\sin(60^o) \approx 0.866$For $R$: $	heta = 45^o$,$2	heta = 90^o$,$\sin(90^o) = 1.0$For $S$: $	heta = 60^o$,$2	heta = 120^o$,$\sin(120^o) = \sin(60^o) \approx 0.866$Comparing the values of $\sin(2	heta)$,the smallest value is $0.5$,which corresponds to body $P$ $(15^o)$.Thus,body $P$ has the shortest range.

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

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