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Final velocity of car

  1. Feb 25, 2004 #1
    OK, many people in my class got a different answer from each other, so could someone verify my work for me please? Here's the problem:

    A car (m=1000kg) driving at 20 m/s goes over a ramp angled at 37 degrees with a height of 5 m. How fast is the car traveling when it hits the ground 15 m below the end of the ramp?

    my work:

    Vyo = 20m/s ( sin37º)
    Vyo = 12.0 m/s

    Vy = sqrt(vyo^2 + 2gy)
    Vy = sqrt ((12m/2)^2 + 2 * 9.8m/s^2 * 15 m)
    Vy = 21 m/s

    So my answer was 21 m/s

    edit: my webspace is down right now so cant post pic

    is this correct or am i doing something wrong?
     
    Last edited by a moderator: Feb 25, 2004
  2. jcsd
  3. Feb 25, 2004 #2
    You forgot something

    When the car hits the ground, it is not travelling perfectly vertical. Remember, for projectile motion, the horizonal velocity remains constant. Therefore you must add the final vertical velocity and the final horizontal velocity together (note they are vectors) to get a resultant velocity, which is on an angle that you must also calculate based on these velocities.
     
  4. Feb 25, 2004 #3
    21 m/s is the velocity in the y direction. The resultant velocity vector (the velocity at which the car hits the ground) would be the sum of the vectors in the x and y directions.

    The resultant vector would be: 12i -21j
    The car would hit the ground at sqrt ( 21^2 + 12^2) m/s

    The direction is given by arctan (21/12) S of E
     
  5. Feb 25, 2004 #4
    oohh, i see, thanks
     
  6. Feb 25, 2004 #5
    one question, could i use the formula:
    sqrt ((.5Vo^2+gho-ghf)/.5) ?

    what i mean:

    sqrt ((.5 (20m/s)^2 + 9.8m/s^2 * 10m)/.5) = 24.4 m/s

    Or would that not be right?
     
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