Question

A projectile is aimed at a mark on a horizontal plane through the point of projection and calls 6 m short when its elevation angle is ${30}^0$. The projectile overshoots the mark by 9 m when the angle of projection is ${45}^0$. The correct distance of the mark is

• A. 30m
• B.
• C.
• D.

Answer 1

The correct option is B

For $\theta$ = ${30}^0$ , range = $R_1$
For $\theta$ = ${40}^0$, range =$R_2$, R – correct distance
First case: $R-R_1$ = 6 $\therefore$ $R=R_1+6$ = $\frac{u^{2}sin60}{g}+6\dots \dots \left(1\right)$
similarly R = $\frac{u^{2}sin90}{g}-9\dots \dots \left(2\right)$ From (1) and (2) R = 51 + 30 $\sqrt{3}$ m