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Projectile Motion of Car Problem

  1. Dec 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24 degrees below the horizontal. The parking break is defective, so the car rolls from rest down the incline with a constant acceleration on 4m/s2 and travels 50m to the edge of the ocean. The cliff is 30 m above the ocean.

    a. What is the car's horizontal position relative to the base of the cliff when it lands in the ocean?

    b. How long is the car in the air?

    2. Relevant equations

    Vy2 = -2gΔy
    Δy = -1/2g(Δt)2
    Vy= -gΔt
    Vx = Δx/Δt

    3. The attempt at a solution

    Δt = square root (30sin(24)/(1/2)g) = 2.5s

    However, I don't really think that's right. If someone could explain the process of solving the overall of equation, that would be extremely helpful.

    ...finals tomorrow.
  2. jcsd
  3. Dec 14, 2009 #2


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    This problem is a bit complicated. First you must calculate the velocity of the car when it reaches the bottom of the incline just before it leaves the cliff. From that velocity vector which makes an angle of 24 degrees below the horizontal, you need to calculate the horizontal and vertical components of its velocity. Call these [itex] V_0x [/itex] and [itex] V_0y [/itex] respectively. Now take your coordinate axes to have their origin at the edge of the cliff where the car leaves. Choose the Y axis vertical, X axis horizontal. Then you need to simultaneously solve the kinematic equations of projectile motion:

    [tex] X = X_0 + V_0x t + \frac{1}{2} a_x t^2 [/tex]

    [tex] Y = Y_0 + V_0y t + \frac{1}{2} a_y t^2 [/tex]

    Recall that [itex] a_y = g [/itex]. What is [itex] a_x [/itex] ?
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