Projectile motion problem?

Question

Please help me.

A projectile is fired in such a way that its horizontal range is equal to three times its maxium height. What is the angle of projection?

I understand that we use the equation R = (V^2 / g) sin2(theta)

What i dont understand is what the horizontal and vertical distances should be.

Am i supposed to use the above equation? or should i use:

x = Vcos(theta)t
-h= Vsin(theta)t - .5gt^2

Thank you!

Answer 1

If you combine the vertical and horizontal equations that you listed by setting them equal to time and then setting them equal to each other thus eliminating time, you get the range equation that is listed above.

After working for a while.....I got 53.1 degrees as the angle that will give a range 3 times the height.

I set vf^2 = vi^2 + 2ad equal to the range equation (set vf = o, apex)
this eliminates velocity and you get... tan (theta) = 4/3

Answer 2

What difference does the apex or range have to do with it - - You're interest in the angle of trajectory. Try solving for that.