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Kinematics help

  1. Jan 31, 2007 #1
    1. The problem statement, all variables and given/known data

    A stunt man drives a car at a speed of 19.3 m/s off a 32.8 m high cliff. The road leading to the cliff is inclined upward at an angle of 19.3°. How far from the base of the cliff does the car land?
    What is the car's impact speed?
    What is the angle of impact?

    2. Relevant equations

    3. The attempt at a solution
    First i divided my initial speed into components so i got Vx=6.38m/s and Vy=18.22m/s

    I started with the y-direction:
    d= v1t+0.5at^2 <---- i solved for time using inital speed at 18.22m/s
    I got t= 1.86s

    I then moved to the x-direction:
    I used v= d/t with v= 6.38m/s
    I got d= 11.87m

    This answer is wrong however i dont know where i went wrong and what i should do to correct it. I didnt get a chance to move on to the other questions either. Any help would be appreciated.
  2. jcsd
  3. Jan 31, 2007 #2


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    Homework Helper

    You mixed up the components of the initial velocity.
  4. Jan 31, 2007 #3
    Welcome to PF, saralsaigh.

    I think you may have the velocity components mixed up.
  5. Jan 31, 2007 #4

    oops yea i see that now, thanks alot:blushing:
  6. Jan 31, 2007 #5

    ok so i switched my speed components around but i seem to still get a very close answer i got dx=11.86m which is wrong also.... :grumpy: HELP!!!
  7. Feb 1, 2007 #6
    Can you show how you worked out this equation d= v1t+0.5at^2 ?
  8. Feb 1, 2007 #7

    ok this is how i worked it out , i started with the y-direction and v1=6.38m/s, d=32.8, and a=-9.8 i subbed those into the equation and i got
    4.9t^2 - 6.38t + 32.8 = 0 so i used the quadratic equation and i got t= 0.651seconds.
    Then i used this time and plugged it into my x-direction and used the formula d=vt with my speed in the x-direction being 18.22m/s and i got an answer for dx= 11.86m.
    Clearly i went wrong somewhere but i cannot point out my error...
  9. Feb 1, 2007 #8
    4.9t^2 - 6.38t + 32.8 = 0

    I worked it out only now, and that equation does not have a solution in the reals.

    Are you sure you typed the question correctly?
  10. Feb 1, 2007 #9
    i think so, idk i used my calculator to get the answer, but heres how i set it up:

    d=32.8m, v1= 6.38m/s (maybe this number should b negative since speed is down in the y-direction, wat do u think?) and a= -9.8m/s^2

    so d=v1 t + .5 a t^2
    32.8 = 6.38t + -4.9 t^2
    4.9t^2 - 6.38t + 32.8 = 0
    then i used my calculator and it gave me t= 0.651s.
  11. Feb 1, 2007 #10
    You've taken the positive y-direction to point up, therefore the y-component of the initial velocity is positive. It is the acceleration that points down, and hence negative.
  12. Feb 1, 2007 #11
    yay i got the right answer, i was doing it wrong in terms of calculating time, i had to first calculate the time it took it to reach its max hight and then calculate the time it took from its max hight to reach the floor.
    so for my y-direction my v1=6.38m/s, and v2= 0 and i know that a= -9.8m/s^2 so i used the equation a= (v2-v1)/t to calculate time and i got t=0.65s and then used the same information in this equation= (v1)^2 - (v2)^2 = 2ad to calculate distance and i got d= 2.08. so i added this distance to my given 1distance of 32.8 and used the equation d= 1/2 a t^2 to find out the time from my max hight to the floor and i found t=2.67. Therefore my total time is 3.32s. I used this time in my x-direction and i found out that dx = 60.49m which is the correct answer.

    Thanks a lot for ur help guys :!!)
  13. Feb 1, 2007 #12
    Good work! :smile:
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