How to get this 2 dimensional equation

[superyoo]

Homework Statement

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

Homework 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.

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

 

[quantumdude]

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.

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

 

[superyoo]

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

 

[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.

 

[quantumdude]

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.