Question

The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is x. The angle of projection with the horizontal is

• A. ${\sin }^{-1}x$
• B. ${\cos }^{-1}x$
• C. ${\sin }^{-1}\left(1/x\right)$
• D. ${\cos }^{-1}\left(1/x\right)$

Answer 1

The correct option is D ${\cos }^{-1}\left(1/x\right)$

let $u$ be the speed and $\theta$ be the angle of of projection

$u_x=ucos\theta$

$u_y=usin\theta$

at max height :

$v_y=0,v_x=ucos\theta$

$x=\frac{u}{v_x}$

$x=\frac{u}{ucos\theta }$

$\frac{1}{x}=cos\theta$

$\theta =co{s}^{-1}\left(\frac{1}{x}\right)$