A projectile is fired with speed v0 at an angle θ from the horizontal as shown in the figure.
Consider your advice to an artillery officer who has the following problem. From his current position, he must shoot over a hill of height H at a target on the other side, which has the same elevation as his gun. He knows from his accurate map both the bearing and the distance R to the target and also that the hill is halfway to the target. To shoot as accurately as possible, he wants the projectile to just barely pass above the hill.
What is the initial speed?
Express v0 in terms of g , R , and H .
Find tg , the flight time of the projectile.
Express the flight time in terms of H and g .

I've already found that
H=[tex]\frac{(v0sin(theta))^{2} }{2g}[/tex]
R=[tex]\frac{v0^{2}sin(2theta) }{g}[/tex]