Show that to hit a target h metres above what was its maximum range position on
a horizontal plane, the initial speed of a projectile projected at the same angle as before, must be increased from V to $\displaystyle \frac{V^2}{\sqrt[]{V^2 - gh}} $ m/s (air resistance is neglected.)


I had the equations of motion as
x = $\displaystyle Vtcos\Theta $
y = $\displaystyle \frac{-gt^2}{2} + Vtsin\Theta $

Putting y = 0
t = 0 or $\displaystyle t = \frac{2Vsin\Theta}{g}$
thus max range is $\displaystyle x = \frac{2V^2sin\Thetacos\Theta}{g}$

new projectile must hit $\displaystyle ( \frac{2V^2sin\Thetacos\Theta}{g}, h)$

what shall I do next?

I gather that my new speed must have theta in it. however, the question does not have a theta. There is nothing I can do that eliminates the theta. Is the question wrong?