How to get this 2 dimensional equation

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AI Thread Summary
The discussion revolves around deriving the range R of a projectile in terms of its maximum height h, specifically showing that R = 4hcotθ and that h = R/4 when the range is maximized. Participants suggest simplifying the range equation by using the identity sin(2θ) = 2sin(θ)cos(θ) and substituting variables to express height in terms of time and initial velocity. One user outlines a method to find the maximum height by determining the time at which the projectile reaches half its range and substituting this into the height equation. The conversation emphasizes the importance of careful substitutions and algebraic manipulation to arrive at the desired results. The thread highlights the collaborative effort to solve a complex physics problem before the impending deadline.
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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
 
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superyoo said:

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.
 
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.
 
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.
 
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. :biggrin:
 
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