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  \frac{V^2}{\sqrt[]{V^2 - gh}} m/s (air resistance is neglected.)

Thanks

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

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

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

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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?