Question

A projectile is launched from a height $h$ making an angle $\theta$ with the horizontal with speed $v_0$. The horizontal distance covered by it before striking the ground $X=v_0\cos \theta \left(\frac{2v_0\sin \theta +\sqrt{x{v}_0^2\sin {\theta }^2-2hg}}{g}\right)$. Find $x$.

Answer 1

in vertical direction (taking upward positive)
$-h=v_0\sin \theta t-\frac{g{t}^2}{2}$
$t^2-\frac{2v_0\sin \theta t}{g}+\frac{2h}{g}$
$t=\frac{2v_0\sin \theta +\sqrt{2v_0^2\sin {\theta }^2-2hg}}{g}$
Now in horizontal direction
$X=v_0\cos \theta \left(\frac{2v_0\sin \theta +\sqrt{2v_0^2\sin {\theta }^2-2hg}}{g}\right)$