# A 'simple' projectile problem.

1. Sep 20, 2005

### kuakkgom

This problem has been giving me headaches for the past two weeks. Here it is, as well as how far I have gotten on it:

A projectile is fired with an initial velocity of 53m/s. Find the angle of projection such that the maximum height is equal to it's range.

So that basically boils down to:

H = R
Vo = 53m/s
ay = -9.81m/s^2
t = unknown
x (angle) = find

According to the nifty sheet the teacher tossed out the applicable formulas would be:

R=Vo^2*sin(2x)/g (since Ho and Hf will be the same)
H=Voy^2/2g.

Making H=R gets the formula Voy^2=2*Vo^2*sin(2x) ...which doesn't help me since x is what needs to be found, and without x one can't break Vo into component form.

Do I have the right idea or have I flubbed it already?

Thanks,

-K

2. Sep 20, 2005

Voy=vo*sinx