Question

A projectile is fired at an angle of ${45}^{\circ }$ with the horizontal. What will be the elevation angle of the projectile at its highest point as seen from the point of projection?

Answer 1

$tan\theta =\frac{H}{\frac{R}{2}}=\frac{2H}{R}=\frac{2.\left(\frac{v_0^{2}si{n}^2{\theta }_0}{2g}\right)}{\left(\frac{2v_{0}cos{\theta }_0.v_{0}sin{\theta }_0}{g}\right)}=\frac{sin{\theta }_0}{2cos{\theta }_0}=\frac{tan{\theta }_0}{2}$
$=\frac{tan{45}^{\circ }}{2}=\frac{1}{2}$
$\therefore \theta =ta{n}^{-1}\left(\frac{1}{2}\right)$