1. ## Projectile motion question

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

2. The horizontal range can be determined by finding the times at which the vertical dispacement is zero.

$\displaystyle v \sin(\theta) \cdot t - \frac{1}{2} \cdot g \cdot t^{2} = 0$

This has two solutions:

t = 0 -- That's when you fired the projectile.

$\displaystyle t_{f} = \frac{2 \cdot v \cdot \sin(\theta)}{g}$ -- This is when it lands.

$\displaystyle t_{m} = \frac{v \cdot \sin(\theta)}{g}$ -- This is the middle of the trip. Perhaps this is the moment of greatest height?

Now, what was the question?

Horizontal Range: $\displaystyle v \cdot \cos(\theta) \cdot t_{f}$

3x Max Height: $\displaystyle 3 \cdot \left(v \sin(\theta) \cdot t_{m} - \frac{1}{2} \cdot g \cdot t_{m}^{2}\right)$

You're almost done. What's next?

3. Cool thanks TKHunny now I've equated those as so:

vcosθ . (2vsinθ)/g = 3[vsinθ . (vsinθ)/g - 1/2g . (vsinθ)/g]

and managed to solve that down to

4vcosθ = 6vsinθ - 3g

but now I'm stuck because I've still got more than one variable. Do I need to find v in terms of θ or something? Sorry about the unreadable equations as well I haven't figured out how to use latex yet

4. My mistake, I forgot to square the t...all fixed now. I eventually got it down to

tanθ = 4/3
θ = 53.13°