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Projectile motion on a slope

  1. Sep 1, 2009 #1
    1. The problem statement, all variables and given/known data
    A projectile has to hit a target located on the hill slope. A line of the slope is described by equation Y=.4bX. A projectile is launched at an angle of 60° with respect to the horizantal. The initial distance between the projectile and the target is 60m. Find the initial velocity of the projectile.

    2. Relevant equations
    [tex] x = v_o cos60 t [/tex]
    [tex] y = v_o sin60 t - \frac{1}{2} a t^2[/tex]
    [tex] a^2 + b^2 = c^2 [/tex]

    3. The attempt at a solution
    So we would have a triagle, hypotenuse equal to 60, base equal to x and height equal to .4bx.
    So we would have...
    [tex] x = v_o cos60 t [/tex]
    [tex] .4bx = \sqrt {60^2 - x^2} = v_o sin60 t - \frac{1}{2} a t^2[/tex]
    But then what?
     
  2. jcsd
  3. Sep 2, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    In the final equation put t = x*/vo*cos(60)
    So,
    sqrt( 60^2 - x^2) = vo*sin(60)*x/vo*cos(60) - 1/2*g*x^2/(vo*cos60)^2

    In the equation, you have two unknown quantities.
    It may not be possible to solve the problem unless you know the value of b.
     
  4. Sep 2, 2009 #3
    Well, since the slope of the hill is variable then that would mean that the time it takes to reach the target isn't always the same. Therefore your answer should be in terms of b.
    So first we would solve for x in terms of b.

    [tex] x^2 +(.4bx)^2 = 60^2 [/tex]
    [tex] x = \frac{60}{^4\sqrt{1+.16b^2}} [/tex]

    We also need to solve for y in terms of b.

    [tex] y = .4bx [/tex]
    [tex] y = \frac{24b}{^4\sqrt{1+.16b^2}} [/tex]

    So now that we know y and x in terms of b we can figure out V in terms of b.

    Substitute in our x value in terms of b...
    [tex] \frac{60}{^4\sqrt{1+.16b^2}} = v_o cos60t [/tex]
    Solve for t...
    [tex] t = \frac{60}{v_o cos60^4\sqrt{1+.16b^2}} [/tex]

    Substitute our new y value in terms of b...
    [tex] \frac{24b}{^4\sqrt{1+.16b^2}} = v_o sin60t - \frac{1}{2}gt^2 [/tex]
    Substitute in t...
    [tex] \frac{24b}{^4\sqrt{1+.16b^2}} = v_o sin60 (\frac{60}{v_o cos60 ^4\sqrt{1+.16b^2}}) - \frac{1}{2}g(\frac{60}{v_o cos60 ^4\sqrt{1+.16b^2}})^2 [/tex]
    After simplifying we get [tex] v_o [/tex] to equal...
    [tex] v_o = \frac{-1800g}{^4\sqrt{1+.16b^2}(24bcos^260-60sin60cos60)} [/tex]

    There we go. To test this I set b=0. That would mean that it would be a flat line and x=60 and y=0 and it works out to be the right answer. [tex] v_o = 26[/tex]

    BTW, i don't know how to do cubed roots or forth roots so I did ^4\sqrt. How do you actually do them?
     
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