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Question

(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin 2	heta}{g}$.Let $R$ be the distance to the target.For $	heta_1 = 15^{\c

Vedclass · Accepted Answer

$\frac{1}{2} \sin ^{-1}\left(\frac{7}{10}\right)$

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(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin 2	heta}{g}$.Let $R$ be the distance to the target.For $	heta_1 = 15^{\circ}$,the range is $R_1 = R - 10$. So,$R - 10 = \frac{u^2 \sin 30^{\circ}}{g} = \frac{u^2}{2g}$.For $	heta_2 = 45^{\circ}$,the range is $R_2 = R + 15$. So,$R + 15 = \frac{u^2 \sin 90^{\circ}}{g} = \frac{u^2}{g}$.Dividing the two equations: $\frac{R - 10}{R + 15} = \frac{u^2/2g}{u^2/g} = \frac{1}{2}$.$2R - 20 = R + 15 \Rightarrow R = 35 \ m$.Now,$R + 15 = \frac{u^2}{g} \Rightarrow 35 + 15 = \frac{u^2}{g} \Rightarrow \frac{u^2}{g} = 50 \ m$.To hit the target,the range must be $R = 35 \ m$. Thus,$35 = 50 \sin 2	heta$.$\sin 2	heta = \frac{35}{50} = \frac{7}{10}$.$	heta = \frac{1}{2} \sin^{-1}\left(\frac{7}{10}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 $15 \ m$. Then the angle of projection to hit the target is:

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