Physics problem (LON-CAPA)

[LJay]

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:

1	13.98 m
2	14 m
3	14.0 m
4	10/72 m
5	10.72 m
6	5.58 m
7	12.09 m
8	0 m
9	18.6 m
10	19.98 m
11	20 m
12	9.16 m
13	9.7 m
14	9.70 m
15	18.64 m

All wrong
blah

any help would be appreciated, i've got 4 tries left

 

[galactus]

Did you use the equations for projectile motion?.

\(\displaystyle x=v_{0}cos(\alpha)t, \;\ y=h_{0}+v_{0}sin(\alpha)t-\frac{1}{2}gt^{2}\)

Since the rock is thrown BELOW the horizontal, the angle would be \(\displaystyle \alpha=-27.5^{o}\) or \(\displaystyle 332.5^{o}\).

\(\displaystyle v_{0}=8.3 \;\ m/s, \;\ h_{0}=10.4 \;\ m, \;\ g=9.8 \;\ \frac{m}{s^{2}}\)

Enter the given values into y and set it equal to 0. Then, solve for t.

Enter this t value into x to find how far out it landed.

 

[LJay]

yea i was on the right track
i just forgot to set my initial-velocity-y to negative 3.8
i was using positive

thats what happens when you study on no sleep

thanks