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Projectile motion car tires question

  1. Aug 28, 2013 #1
    Hi guys, I am really stuck with this homework question... The answer is 14.5 m/s but I do not know how to calculate it.

    A car has run into a fire hydrant and come to an abrupt stop. A suitcase tied to a rack on top of the car has been thrown off and has landed on the roadside 11.6 m away from the hydrant. The suitcase is found to have been 1.2 m above the ground when it was still on the rack. Determine the impact speed of the car if the launching angle for the suitcase was 10 degrees (from the positive x-axis).

    (Assume g=9.8 m/s2).

    Thanks!
     
  2. jcsd
  3. Aug 28, 2013 #2
    What have you tried so far? Or are you completely lost?
     
  4. Aug 28, 2013 #3

    kreil

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    Gold Member

    The first thing you should always do is draw a picture. Label all of your knowns and unknowns.

    Next, look for the mechanics equations you need to use based on what your unknowns are.
     
  5. Aug 28, 2013 #4
    I have tried all the motion formulas, tried to find time and vertical velocity at maximum, tried rearranging and substituting and other stuff but I never seem to get the right answer. I initially used the equation: v2 = u2 + 2as to find "u" when "v" is 0 by substituting "s" for 1.2 m. I got 4.85m/s as the vertical component but that would not be correct as the vertical component of 14.5 is 2.52 m/s. I have been trying other methods for 2 hours but I have not been successful in finding the answer.
     
  6. Aug 28, 2013 #5
    Perhaps this is a stupid question, but do you have any detail on how the car is moving before the impact?? Because I would say that the velocity of the car is just horizontal, but then it does not make sense that the suitcase leaves with an angle of 10 degrees... unless there is something else, or the car was moving upwards... which would seem strange
     
  7. Aug 28, 2013 #6
    No, the question I posted was a direct quote from my textbook so that is all the information available.
     
  8. Aug 28, 2013 #7

    gneill

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    Staff: Mentor

    The luggage rack could be mounted at an angle on the car, providing a launch slope.
     
  9. Aug 28, 2013 #8

    Zondrina

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    Homework Helper

    The question is missing information. You need the velocity of the suitcase or the velocity of the car or some time.
     
  10. Aug 28, 2013 #9

    BruceW

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    I'm pretty sure there is enough information there. It is essentially a projectile problem.
     
  11. Aug 28, 2013 #10
    Your main problem with this is that your "v" is not zero when it hits the ground it's zero after it hits the ground (and bounces around or whatever) so it cannot be the case for this scenario. What you should do is work equations for both the vertical and horizontal components. s=1/2at^2-u*sin(10) for the vertical and u*cos(10)t=x for the horizontal, you can get both equations to have just the velocity and time in them and then you have two equations and two unknowns. Careful with signs though, I don't think I accounted for them correctly, was just giving you equations to work with.
     
  12. Aug 28, 2013 #11

    CAF123

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    Gold Member

    If you write explicit equations for the time taken for the projectile to a) first reach to 'rack' level again and b) to make it to the ground from 'rack' level, then you will have one equation with one unknown (albeit the eqn looks messy, but still solvable after some algebra).
     
  13. Aug 28, 2013 #12
    Everything you need is there and the answer is correct, too. I checked it.
    You just have to take the motion equations and the second newton law. If you solve your equations correctly you are done.
     
  14. Aug 28, 2013 #13

    Zondrina

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    Homework Helper

    Upon re-reading the question I see this might be useful :

    $$Δd_H = \frac{v_R^2 sin(2θ)}{a}$$

    I didn't notice the launch angle was already given earlier. This will help find the speed that the suitcase was launched.
     
  15. Aug 29, 2013 #14
    Thank you for the replies! I will attempt this question again and hopefully get it right :)
     
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