Physics
posted by Lindsay .
A catapult on a cliff launches a large round rock towards a ship on the ocean below. The rock leaves the catapult from a height H = 35.0 m above sea level, directed at an angle theta = 46.7° above the horizontal, and with a speed v = 28.4 m/s. Assuming that air friction can be neglected, calculate the horizontal distance D traveled by the projectile.
Could someone give me the first step? I know that I will eventually use delta x = vx x t, but what do I need to do before that?

The first step is the vertical equation, to determine the time in air.
Determine the vertical and horizontal components of the initial velocity.
Vertical equation.
Hfinal=Hinitial + Viv*t  1/2 g t^2
solve for t, the time in air. Then work the horizontal equation.
dfinal=vih*t 
Thanks!

Height reached above the cliff is h = V^2sin^2(µ)/9.8.
t1 = Vsin(µ)/9.8.
h + 35 = 9.8t2^2/2>t2
d = Vcos(µ)(t1 + t2)
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