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Basic Mechanics - Projectiles Q's having problems

  1. Dec 2, 2007 #1
    1. The problem statement, all variables and given/known data
    In a film stunt a car is driven over a cliff at 15ms^-1, the ground at the top of the cliff being level. The wreckage is 45m horizontally from the foot of the cliff.

    What is the vertical height that the car has fallen?

    2. Relevant equations
    Im not sure but im guessing s=ut+1/2at^2 should be used here.

    3. The attempt at a solution
    I've attempted to use the equation above, using v=u+at to get a value for t and using v=0, a=-9.8 and u=15. Now after finding a value for t and plugging that into the s=... eqn im not getting the right answer.

    Am i going about this the wrong way? Thanks a lot for your help.
  2. jcsd
  3. Dec 2, 2007 #2
    Use x = x0 + v0t + .5at^2
    The car is not accelerating so all you need is x = v0t. Plug in your values...

    45 = 15t
    t = 3 seconds

    Using x = .5gt^2
    x = .5 * (9.8) * 3^2
    x = 44.1 m
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