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How to get this 2 dimensional equation

  1. Feb 25, 2009 #1
    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. jcsd
  3. Feb 25, 2009 #2

    Tom Mattson

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    Forum rules require you to post an attempt at the solution before receiving help. So let's see what you've got so far.
     
  4. Feb 25, 2009 #3
    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.
     
  5. Feb 25, 2009 #4

    LowlyPion

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    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.
     
  6. Feb 25, 2009 #5

    Tom Mattson

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

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

    [tex]T=\frac{v_0\sin(\theta)}{g}[/tex]

    Don't forget that [itex]\sin(2\theta)=2\sin(\theta)\cos(\theta)[/itex].

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

    Combine this equation for [itex]H[/itex] with the equation for [itex]R[/itex] to obtain the result. I did it very quickly by eliminating [itex]g[/itex] 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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