Question

A projectile is fired at an angle of ${45}^0$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Refer Figure
Max. height , H =$\frac{u^{2}si{n}^2\times {45}^0}{2g}=\frac{u^2}{4g}=AC$

Horizontal range, R = $\frac{u^{2}sin2\times {45}^0}{g}=\frac{u^2}{g}\therefore OC=R/2=\frac{u^2}{2g}Tan\alpha =\frac{AC}{OC}=\frac{u^2/4g}{u^2/2g}=\frac{1}{2}\therefore \alpha =ta{n}^{-1}\left(1/2\right)\alpha =ta{n}^{-1}\left(1/2\right)$