# Catapult on a cliff

1. Jan 30, 2005

### booneyromper

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 of 34.0m above sea level, directed at an angle above the horizontal with an unknown speed. The projectile remains in flight for 6.00 seconds and travels a horizontal distance of 174m. Assuming that air friction can be neglected, calculate the value of the angle (in degrees). Also calculate the speed at which the rock is launched.

My brains are cooked. I'm not sure where to begin with this one.

2. Jan 30, 2005

### Curious3141

Cooked brains or not, please show your reasoning and lots of working.

What are the equations governing the motion of a projectile ?

3. Jan 30, 2005

### booneyromper

Thanks for the hint. I took a break, ate dinner, shook off the frustration, then came back and solved the problem
For horizontal velocity X = Xo + (Vxo)t,
174m = 0+(Vxo)6.00s,
Vxo = 29 m/s
For vertical velocity Y = Yo + (Vyo)T - 1/2gt^2,
-34m = (Vyo)6.00s - (1/2)(9.8m/s^2)(6.00s^2)
Vyo = 23.73m/s
So the angle theta = tan-1 (23.73/29) = 39.29deg
and the rock is launched at... Square root of (29^2 + 23.73^2) = 37.47

The moral of the story is don't do physics on an empty stomach.