# Projectile motion question

• Nov 10th 2009, 06:06 PM
neobeatlemaniac
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 (Happy)
• Nov 10th 2009, 06:38 PM
TKHunny
The horizontal range can be determined by finding the times at which the vertical dispacement is zero.

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

This has two solutions:

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

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

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

Now, what was the question?

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

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

You're almost done. What's next?
• Nov 10th 2009, 07:12 PM
neobeatlemaniac
Cool thanks TKHunny (Happy) 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 (Worried)
• Nov 10th 2009, 07:40 PM
neobeatlemaniac
My mistake, I forgot to square the t...all fixed now. I eventually got it down to

tanθ = 4/3
θ = 53.13°