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(B) Let $d$ be the distance of the target.The horizontal range of a projectile is given by $R = \frac{u^2 \sin 2	heta}{g}$.For $	heta = 15^{\circ}$,

Vedclass · Accepted Answer

$\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)$

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(B) Let $d$ be the distance of the target.The horizontal range of a projectile is given by $R = \frac{u^2 \sin 2	heta}{g}$.For $	heta = 15^{\circ}$, the range is $R_1 = \frac{u^2 \sin(30^{\circ})}{g} = \frac{u^2}{2g}$.Given that it falls short by $10 \ m$, we have $R_1 = d - 10$, so $\frac{u^2}{2g} = d - 10$ (Equation $i$).For $	heta = 45^{\circ}$, the range is $R_2 = \frac{u^2 \sin(90^{\circ})}{g} = \frac{u^2}{g}$.Given that it is away (beyond) by $10 \ m$, we have $R_2 = d + 10$, so $\frac{u^2}{g} = d + 10$ (Equation $ii$).Dividing Equation $i$ by Equation $ii$, we get $\frac{1}{2} = \frac{d - 10}{d + 10}$.Solving for $d$, $d + 10 = 2d - 20$, which gives $d = 30 \ m$.Substituting $d = 30$ into Equation $ii$, we get $\frac{u^2}{g} = 30 + 10 = 40$.To hit the target at $d = 30 \ m$, we set $R = 30 = \frac{u^2 \sin 2	heta}{g} = 40 \sin 2	heta$.Thus, $\sin 2	heta = \frac{30}{40} = \frac{3}{4}$, which implies $	heta = \frac{1}{2} \sin^{-1}\left(\frac{3}{4}ight)$.

$A$ particle aimed at a target, projected with an angle $15^{\circ}$ with the horizontal is short of the target by $10 \ m$. If projected with an angle of $45^{\circ}$ it is away from the target by $10 \ m$, then the angle of projection to hit the target is

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