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Projectile motion problem

  1. Jan 19, 2005 #1
    Problem : find a [tex]v_i[/tex] and angle [tex]\theta[/tex] that will be able to hit the target 795 m away and clear the 60 m hill 600 m away. I attached a picture of the problem.

    I was trying to do this using the equation
    [tex]y=y_0+(x-x_0)*tan(\theta)-\frac {1}{2}g (\frac {x-x_0}{v_0*cos(\theta)})^2[/tex]

    I used this by seperating the problem into two parts from the origin to x=600 m and y=60m ([tex]y_0 = x_0 = 0[/tex]) and the second part from x= 795 m and y=0 m([tex]y_0=60m , x_0=600m[/tex])
    I know this is wrong, because I couldn't get an answer. I was wondering what the best way to approach this problem would be. Thanks in advance!
     

    Attached Files:

    Last edited: Jan 20, 2005
  2. jcsd
  3. Jan 20, 2005 #2

    Andrew Mason

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    Homework Helper

    Try using your equation and finding the expressions for [itex]v_0[/itex] and [itex]\theta[/itex] where (x,y) = (600,60) and (x,y) = (795,0)

    AM
     
  4. Jan 20, 2005 #3
    Hi,
    I tried to do this before, I couldn't solve for either variable and when I plug it into my calculator to solve it says that it's false. This is where I got stuck. Is there another way to solve for it or am I doing something wrong?
    [tex]60=600*tan(\theta) - \frac{1}{2}*g*\frac{600}{v_0*cos(\theta)}
    [/tex]
    and
    [tex] 0=795*tan(\theta)-\frac{1}{2}*g*\frac{795}{v_0*cos(\theta)}
    [/tex]
     
    Last edited: Jan 20, 2005
  5. Jan 20, 2005 #4

    Andrew Mason

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    In the last term, t (= x/vcos) should be squared:

    [tex]60=600*tan(\theta) - \frac{1}{2}*g*\frac{600^2}{v_0^2cos^2(\theta)}[/tex]
    and
    [tex] 0=795*tan(\theta)-\frac{1}{2}*g*\frac{795^2}{(v_0^2cos^2(\theta))}[/tex]

    Substituting [itex]t=x/v_0cos(\theta)[/itex] into the general equation of motion, you get the general solution to the quadratic for the range, R:

    [tex]R = v_0^2sin^2\theta/2g[/tex]

    Then substitute that into the equation of motion to get:
    [tex]y = xtan\theta - \frac{1}{4R}tan^2\theta[/tex] where R = range (795)

    By substituting x,y = 600,60 you have a quadratic equation in terms of tan(theta) only. You will have to check my algebra - I am not guaranteeing that I haven't missed something.

    AM
     
    Last edited: Jan 20, 2005
  6. Jan 20, 2005 #5
    Hi,
    Someone asked my professor about this in class and he said we only need to use the equation [tex]y=y_0+(x-x_0)*tan(\theta)-\frac {1}{2}g (\frac {x-x_0}{v_0*cos(\theta)})^2[/tex] Once.
    He said that we need to put [tex]v_0[/tex] in terms of [tex]\theta[/tex]. and find a [tex]\theta[/tex] that would give a minimum value of [tex]v_0[/tex].
    [tex]v_0=sqareroot((\frac{.5*g*x^2}{x*tan(\theta)(cos^2(\theta))}))[/tex]
    using the (795,0) and [tex]x_0 = y_0 = 0[/tex]
    I was trying this but I dont recall how to find an angle that would give a minimum value. Can you offer any help regarding this ?
    Thanks
     
    Last edited: Jan 20, 2005
  7. Jan 20, 2005 #6
    Thanks for your help, I found the answer!
     
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