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Motion Help Please

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data
    A stunt car is driven from a 10° ramp at 40m/s. How far away should the landing ramp be. (It lands at the same level as it was launched from.)


    2. Relevant equations
    Answer is 54.7m


    3. The attempt at a solution
    v^2 = u^2 + 2as
    (v^2 - u^2)/(2a) = s
    (0 - 40^2) / (2 x 9.8) = -81.6
     
  2. jcsd
  3. Mar 5, 2012 #2
    try splitting the original velocity into x and y components. Remember that gravity only affects the velocity component in the y direction.
     
  4. Mar 5, 2012 #3
    u r right. gravity will affect only vertical component. horizontal component remains same.
     
  5. Mar 5, 2012 #4
    I don't understand. What equations am I suppose to use?
     
  6. Mar 5, 2012 #5
    Well, if the velocity of the stunt car is 40 m/s at 10 degrees initially, then how would you find the x and y components of its velocity at that moment?
     
  7. Mar 5, 2012 #6
    x = 40cos10 = 39.39
    y = 40sin10 = 6.95

    I get that part but I'm not sure how that helps me achieve the conclusion? ><
     
  8. Mar 5, 2012 #7
    Well so if the car flies off into the air, how do you find out how long it's going to stay in the air?
     
  9. Mar 5, 2012 #8
    v = u + at?
    0 = 6.95 +9.8t
    -6.95 / 9.8 = t
    0.7 = t
    0.7s
     
  10. Mar 5, 2012 #9
    that's what I got. So now you know how long the car is in the air and how fast it is going in the x direction, how would you find out how far away it lands?
     
  11. Mar 5, 2012 #10
    .7 x 39.39
    = 27.573
    Then do I multiply this number by 2?
     
  12. Mar 5, 2012 #11
    well, what do you think? ;)
     
  13. Mar 5, 2012 #12
    got it, cheers buddy!!
     
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