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Projectile Motion: Find Launch Speed

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Your alpine rescue team is using a slingshot to send an emergency medical packet to climbers stranded on a ledge on the side of a mountain. If the height of the ledge is 270m (y = 270m) and the ledge is 390m from the slingshot (x = 390m), what is the initial velocity of the slingshot if the launch angle is 70 degrees?

    V(o) = ?
    x = 390m
    y = 270m
    Theta = 70 degrees
    a(y) = -9.81
    a(x) = 0

    2. Relevant equations

    Since the only acceleration is constant, I know that I must solve this problem using these equations:

    V = V(0) + at
    X = X(0) + V(0)t + 1/2at^2
    V^2 = V(0)^2 + 2ax

    3. The attempt at a solution

    When I first attempted the problem, I figured I would have to solve for variables using the y-axis.

    I used a variation of the third equation:

    V(y)^2 = V(0)(y) + 2ay

    And solved for V(0)(y). (V(y) is 0 when it hits the ledge):

    0 = V(0)(y) + 2(9.81)(270)

    I got:

    72.78 = V(0)(y)

    and then I plugged it into:

    sin(70) = V(0)(y) / V(0)

    V(0) = V(0)(y) / sin(70)

    V(0) = 77.45 m/s

    This was the only valid approach I could think of, but the online HW program I'm using says that this is the incorrect answer. I'm sure that this is very simple and I am just ignorant of something, but can anyone provide any pointers to help me out?

    Thanks!
     
  2. jcsd
  3. Sep 17, 2011 #2

    Delphi51

    User Avatar
    Homework Helper

    I didn't see that given in the question. It would be interesting to use the basic distance equations to see if your velocity causes the package to hit the ledge. Spreadsheet?

    The two equations define the trajectory. You just need to put in the (390,270) to make the hit. Write the horizontal distance equation 390 = Vi*cos(70)*t
    and the vertical distance equation 270 = Viy*t + ½*a*t².
    That gives you two unknowns (t and Vi) and two equations.
    They solve to Vi somewhat larger than you got.
     
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