Question

A projectile is aimed at a mark which is in the same horizontal plane as the point of projection. It falls $10m$ short of the target when it is projected with an elevation of ${75}^{\circ }$ and falls $10m$ ahead of the target when it is projected with an elevation of ${45}^{\circ }$. Find the correct elevation of projection so that it exactly hit the target. It is given that the initial velocity of projection is the same in each case.

• A. $\frac{1}{2}{\sin }^{-1}\frac{3}{4}$
• B. $\frac{1}{2}{\cos }^{-1}\frac{4}{5}$
• C. $\frac{1}{2}{\sin }^{-1}\left(\frac{4}{5}\right)$
• D. $\frac{1}{2}{\sec }^{-1}\left(\frac{5}{4}\right)$

Answer 1

The correct option is A $\frac{1}{2}{\sin }^{-1}\frac{3}{4}$

$\begin{array}{c}\frac{v^2}{g}=d+10\\ \frac{v^2\sin \left(2\times {75}^{\circ }\right)}{g}=d-10\\ \Rightarrow \frac{v^2}{2g}=d-10\\ \Rightarrow \frac{v^2}{g}+\frac{v^2}{2g}=2d\\ \Rightarrow \frac{3v^2}{2g}=2d\\ \Rightarrow \frac{v^2}{g}\left(\frac{3}{4}\right)=d\\ \sin 2\theta =\frac{3}{4}\\ \therefore \theta =\frac{1}{2}{\sin }^{-1}\left(\frac{3}{4}\right)\end{array}$

Hence, the option $A$ is correct answer.