A projectile is launched with speed $\displaystyle V_o$ at an angle $\displaystyle \alpha_{o}$ above the horizontal. The launch point is at a height h above the ground.

Show that if air resistance is neglected, the horizontal distance that the projectile travels before striking the ground is:

$\displaystyle x = \frac{V_{o}\cos{\alpha_{o}}}{g}(V_{0}\sin{\alpha_{ o}} + \sqrt{V_{o}^2\sin^2{\alpha_{o}} + 2gh})$

Verify that if the launch point is at ground level so that h = 0 this is equal to the horizontal range R = $\displaystyle \frac{{V_{0}}^2\sin{2\alpha_{o}}}{g}$ (derived from the example in my texbook)

Ok.....

the projectile is the first case and the second case is the angle below h ...

and im very confused at which equations to gather here:

$\displaystyle x = V_{o}\cos{\alpha_o}

t$

$\displaystyle y = (V_{o}\sin{\alpha_o})t - \frac{1}{2}gt^2 $

$\displaystyle V_x = V_o\cos{\alpha_o}$

$\displaystyle V_y = V_o\sin{\alpha_o}$

alpha_o is initial angle

help needed for physics pros here (Especially topsquark)