Answer
$\frac{v_{\circ}^2tan^2\theta}{2g}$
Work Step by Step
We can find the required height difference as follows:
$(v_{fy})^2=(v_{\circ y})^2-2g(-h)$
$\implies (-v_fsin\theta)^2=(0)^2+2gh$
This simplifies to:
$2gh=v_f^2sin^2\theta$....eq(1)
As $v_{\circ x}=v_{fx}$
$\implies v_{\circ}=v_fcos\theta$
$\implies v_f=\frac{v_{\circ}}{cos\theta}$
We plug in this value of $v_f$ in eq(1) to obtain:
$\implies 2hg=(\frac{v_{\circ}}{cos\theta})^2sin^2\theta$
This simplifies to:
$h=\frac{v_{\circ}^2tan^2\theta}{2g}$
