Physics Problem (mothereffing loncapa)

The defender of a castle throws a rock off of the castle wall, releasing the rock at a point 10.4 m above the ground. The rock is thrown at 8.3 m/s at an angle of 27.5 ° below horizontal. What is the horizontal distance from the wall to the point where the rock hits the ground?

I've gotten:

13.98 m
14 m
14.0 m
10/72 m
10.72 m
5.58 m
12.09 m
0 m
18.6 m
19.98 m
20 m
9.16 m
9.7 m
9.70 m
18.64 m

plz help a dummy out
got 4 tries left

First we need to calculate just the vertical component of the initial speed:

sin(27.5 degrees) = (vertical speed)/8.3

vertical speed = 3.8m/s downwards

We then use the formula, where we replace the initial speed with the vertical speed:

Height = (initial speed)*(time) + 0.5*(acceleration)*(time^2)

So,

10.4m = (3.8m/s)*(time) + 0.5*(9.81m/s^2)*(time^2)

Using a calculator to solve for time gives:

time = 1.12 seconds

Which tells us how long the rock is in the air before it hits the ground.
We now calculate the initial horizontal speed of the rock:

cos(27.5) = horizontal speed/8.3

horizontal speed = 7.4m/s

This value multiplied by the time gives us the horizontal distance the rock has travelled (as the horizontal speed is assumed constant throughout).

So, horizontal distance = (7.4m/s)*(1.12s) = 8.2m


Hope that helped.