# How to get this 2 dimensional equation

1. Feb 25, 2009

### superyoo

1. The problem statement, all variables and given/known data
Show that the range R can be expressed in terms of the maximum height h, and in particular that R = 4hcotθ. (b) Show that, when the range is a maximum, h = R/4

2. Relevant equations
R = ( V0^2sin(2θ) ) / g
V0^2 = ( (1/4)g^2t^4+R^2 ) / t^2
Vy = Vy0 - gt

R represents displacement or range, v0 represents the initial velocity, g represents gravity, t is time and vy0 represents the y component.

3. The attempt at a solution
I tried approximately for 4 hours and still couldn't get it. It would be very nice if someone helps me.

Thanks a lot.

ps Its due tomorrow

2. Feb 25, 2009

### Tom Mattson

Staff Emeritus
Forum rules require you to post an attempt at the solution before receiving help. So let's see what you've got so far.

3. Feb 25, 2009

### superyoo

I tried many different times.
My last try :
http://www.webforone.com/images/ko1w9b5lze83cqfl4ujx_thumb.png

I didn't scan all my tries because I want to make sure that someone in this website knows the answer for this question, before I spend my time in scanning my work.

4. Feb 25, 2009

### LowlyPion

Welcome to PF.

First of all make your Range equation less complicated and undo the sin2θ to 2sinθcosθ.

Now I see you derived an expression for H in terms of Vo*sinθ - eliminating t from the equation

h = 1/2*g*t2

So look carefully at your expanded range equation and see how you might make a substitution.

5. Feb 25, 2009

### Tom Mattson

Staff Emeritus
Thanks for posting your work. After looking it over, I have a better way to do this.

1.) The maximum height occurs when $x=\frac{R}{2}$ (that is when the projectile has covered half of its range). Set $x(t)=v_0\cos(\theta)t$ equal to $\frac{R}{2}$ and solve for $T$. This is the time at which the particle reaches its maximum height. You should be able to show that:

$$T=\frac{v_0\sin(\theta)}{g}$$

Don't forget that $\sin(2\theta)=2\sin(\theta)\cos(\theta)$.

2.) Plug this $T$ into the equation for $y(t)=v_0\sin(\theta)t-\frac{1}{2}gt^2$. This will give you the maximum height $H$.

Combine this equation for $H$ with the equation for $R$ to obtain the result. I did it very quickly by eliminating $g$ in the first one and plugging into the second one.

Follow my steps and you'll have this done in plenty of time to catch South Park.