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Projectile motion car acceleration problem

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data

    A car is parked near a cliff overlooking the ocean on an incline that makes an angle of 29.3 degrees with the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline and has a velocity of 9m/s when it reaches the edge of the cliff. The cliff is 15.7m above the ocean. The acceleration of gravity is 9.8m/s^2

    2. Relevant equations

    [itex]\theta[/itex] = 29.3
    Velocity = V = 9 m/s
    Initial height = 15.7 m
    V_x = V cos [itex]/theta[/itex]
    V_y = V sin[itex]/theta[/itex]
    r_y = yo +V_oyt + 1/2gt^2 //equation for y component of the motion
    r_x = V_xt //equation of x component of the motion

    3. The attempt at a solution
    1. Find x component
    7.848263 = 9 cos (29.3)

    2. Find y component
    4.404444 = 9 Sin ( 29.3)

    3. Find When Y motion = zero

    0 = 15.7 -4.404444t - 1/2(9.8)t^2
    t = 1.424256 //The car hits the water at this time

    4. See how far the car goes in the X direction during the same time interval
    X = 11.178448

    Attached Files:

    Last edited: Sep 16, 2011
  2. jcsd
  3. Sep 16, 2011 #2


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    The y-component of the velocity and the acceleration of gravity must have the same sign because they are both downward.
  4. Sep 16, 2011 #3


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    I got 11.0 meters. Probably rounding differences. I used,

    0 = 15.7 - 4.4*t - 4.9*t^2

    and solved for t.
  5. Sep 16, 2011 #4
    I did the actual calculation with -9.8m/s^2. I made a typo in the post. Fixed it now.
  6. Sep 16, 2011 #5
    I took it out to 6 decimal units because that is what my HW system requires. I am still getting the wrong answer every time. Am I missing something?
  7. Sep 16, 2011 #6
    I emailed my professor and he responded with " You are solving a quadratic equation to find the time the car takes to hit the water, which is a recipe for wrong arithmetic. You can find the time using a first-order equation, and this is by far the best way to do it... much less chance of making a mistake."

    What equation will let me know when height = 0 when i have the velocity?
  8. Sep 16, 2011 #7


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    I solved the quadratic and I got 10.957633 m. It's a ridiculous number of decimal places, but that's what it is. Probably your professor is referring to the following derivation for the time.

    1. Start with vy2 = v0y2 + 2aΔy
    2. Solve for vy by taking the square root.
    3. Substitute the result in vy=v0y + at
    4. Solve the resulting equation for the time.

    I am not convinced that this four-step derivation provides "much less chance for making a mistake" than the one-step solution of the quadratic. Take your pick.
  9. Sep 17, 2011 #8
    Thanks for the help guys. I finally got it. U guys are the best!
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