Posted by **Lindsay** on Tuesday, November 6, 2007 at 3:46pm.

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?

- Physics -
**bobpursley**, Tuesday, November 6, 2007 at 3:52pm
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

- Physics -
**Lindsay**, Tuesday, November 6, 2007 at 4:02pm
Thanks!

- Physics -
**tchrwill**, Tuesday, November 6, 2007 at 4:41pm
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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