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Homework Help: Film stunt physics car problem

  1. Nov 9, 2005 #1
    In a film stunt, a car is driven off a cliff with a horizontal velocity of 12.9 ms-1. The cliff face is vertical, and the cliff is 20.7 m high. Calculate the distance in m from the base of the cliff to the point where the car strikes the ground.

    should i be considering this problem in x and y - components of the motion separately.

    taking upwards as +ve direction for vectors
    displacment = -20.7
    initial displacment = 0
    inital speed = 12.9
    a=-9.8

    using it in s = ut + 1/2at^2 i get
    20.7 + 12.9t - 4.9t^2
    t = 3.757 +ve value taken

    s=ut
    12.9*3.757= 48.5m as the displacement
    and then 48.5 + 20.7 = 69.2m for the distance.
    help would be apprecated
    thanks in advance
     
  2. jcsd
  3. Nov 9, 2005 #2

    BobG

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    Science Advisor
    Homework Helper

    Only one mistake.

    The horizontal velocity is 12.9 m/sec. There is no initial vertical velocity.

    But, your overall approach of solving the x component and y component separately is right. Solve the y component to find the time. Use that time to solve the horizontal component.
     
  4. Nov 9, 2005 #3
    cheers!!!!!!!!!!!!!!!!!
     
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