- #1

alfredo24pr

- 49

- 0

## Homework Statement

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 34.0 m above sea level, directed at an angle θ above the horizontal with an unknown speed v0.

The projectile remains in flight for 6.00 seconds and travels a horizontal distance D of 149.0 m. Assuming that air friction can be neglected, calculate the value of the angle θ.

## Homework Equations

y=y0 + v0yt - 0.5gt^2

tan θ = y/x

## The Attempt at a Solution

I used y=y0 + v0yt - 0.5gt^2

y= -34 + 0 + 0.5 (9.8) (6^2)

y= -34 + 4.9(36)

y= -34 + 176.4

y= 142.4m

For x, the distance is given: 149 m

The angle:

tan θ = y/x

tan θ = 142.4/149

θ = 43.7

Is that correct? LONCAPA does not accept my answer, so if it is right I will assume that it is the degrees unit.