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AP EAMCET · PHYSICS · Motion In Two Dimensions

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

  1. A \(\frac{1}{2} \sin ^{-1}\left(\frac{1}{10}\right)\)
  2. B \(\frac{1}{2} \sin ^{-1}\left(\frac{3}{10}\right)\)
  3. C \(\frac{1}{2} \sin ^{-1}\left(\frac{9}{10}\right)\)
  4. D \(\frac{1}{2} \sin ^{-1}\left(\frac{7}{10}\right)\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

As Range \(\propto \sin 2 \theta\) So, \(\frac{R-10}{R+15}=\frac{\sin 2 \theta_1}{\sin 2 \theta_2}=\frac{\sin 30^{\circ}}{\sin 90^{\circ}}=\frac{1}{2}\) \(\Rightarrow \quad 2 R-20=R+15 \Rightarrow R=35\) For the range to be maximum \(\sin 2 \theta=\sin 90^{\circ}=1\) So,…
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