Hi. I'm having trouble solving this problem:

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

I tried using the formulas for x and y displacement as so:

x-displacement = vcosθt

y-displacement = vsinθt - 1/2gt^2

therefore

vcosθt = 3(vsinθt - 1/4gt^2) where time = 1/2t because this is when the projectile is at its highest point

But I can't get rid of the t's in the equation so I think I'm going about this the wrong way. Could somebody point me in the right direction? Thanks