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Easy question on exam that I couldnt figure out HELP

  1. Dec 22, 2006 #1
    A boy throws a rock horizontally 12 m/s at a pigeon on a fence 5m away. Unfortunatley the rock the missed. By how much did the rock miss?

    I ve got my kinematic stuff down pat, but I just had no idea how to go about this problem.

    thanks for any help.
     
  2. jcsd
  3. Dec 22, 2006 #2

    Doc Al

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    Assume that the boy aimed directly at the pigeon.
     
  4. Dec 22, 2006 #3

    berkeman

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    ...and how long did it take the rock to hit the fence?
     
  5. Dec 23, 2006 #4
    the time was not given, that why I had trouble with it. All the information above is exactly what it said on the test.
     
  6. Dec 23, 2006 #5

    Doc Al

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    You should be able to figure out the time it takes for the rock to hit the fence--you are given all the information needed.
     
  7. Dec 23, 2006 #6
    assuming g=10,i get it as ~0.5 m....
     
  8. Dec 23, 2006 #7
    Assuming the boy has sufficient height ie, the stone strikes the fence , all the information required is given in the question.
    However your answer seems to be incorrect. Post your attempt if you can and we'll try to correct you.
     
  9. Dec 23, 2006 #8

    Doc Al

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    Please give Aristodol the opportunity to solve his own problem. (And if you wish to help, don't just toss out answers.)
     
  10. Dec 23, 2006 #9
    I no that assuming that x is 5 you get a time .42s i believe. but thats not the time we actually need so that is where i got stuck. I sat there for a long time pondering this question.
     
  11. Dec 23, 2006 #10

    Doc Al

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    That is most definitely the time you need. Keep going.
     
  12. Dec 23, 2006 #11
    Ive been trying to solve this problem for severall days I have been unable to do it. If you could please just show me the way to go about it it would be appreciated. Ive tried many of time but i keep getting stuck.
     
  13. Dec 23, 2006 #12

    Doc Al

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    Explain what you've been doing and where you are getting stuck. You have correctly calculated the time it takes for the rock to reach the fence.

    Hint: If there were no gravity, the rock would go in a straight line and hit the pigeon. But gravity acts. During the time it takes the rock to get to the fence, how far does it fall?
     
  14. Dec 23, 2006 #13
    I have established that t=.42s we no that Vx=Vox=12m/s also Vfy=0m/s ay=-9.8m/s O i see it now

    Vf=Vo + at---> 0=Voy + -9.8(.42) Voy= 4.12m/s
    now whats throwing me off is that i have the velocities for x and y and im given a distance in 5 but thats not the correct distance
     
  15. Dec 23, 2006 #14

    Doc Al

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    Good. You have the correct initial velocity, both horizontal and vertical components.

    You have found the vertical component of the rock's speed at the moment it hits the fence. But what you need is the vertical distance it travels in that time.
    5 m is the horizontal distance it travels; you've already used it to find the time of travel.

    What kinematic equation relates distance to time for constant acceleration?
     
  16. Dec 23, 2006 #15
    y=4.12(.42) + 1/2 (-9.8)(.42)^2
    1.73 + -.86
    y=.87m ok???
     
  17. Dec 23, 2006 #16

    Doc Al

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    Not exactly. The equation you are using (for vertical displacement) is this:
    [tex]y = v_0 t + (1/2) a t^2[/tex]

    But what is the vertical component of the initial velocity?
     
  18. Dec 23, 2006 #17
    huh? not exactly sure what u mean by that?
     
  19. Dec 26, 2006 #18

    berkeman

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    In an earlier post of yours, you correctly identified the initial vertical velocity of the rock. Doc is asking where you came up with the 4.12 term. What is it? It certainly is not the initial vertical velocity of the rock.

    BTW, I think that the final answer you will get is for how far below the bird the rock hits, correct?
     
  20. Dec 26, 2006 #19
    4.12 is indeed the initial vertical velocity I solved for it and he agreed that that was indeed the correct number.

    Vf=Vo + at---> 0=Voy + -9.8(.42) Voy= 4.12m/s
     
  21. Dec 27, 2006 #20

    Doc Al

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    Nope. 4.12 m/s is not the correct initial vertical speed! 4.12 m/s downward is the vertical component of the velocity when the rock hits the fence--which is not needed to solve this problem.

    I probably misread this before. For some reason, you have set Vf = 0. Why? (I had assumed that you were calculating Vf, not Vo. My bad. :redface: )

    In any case, the initial vertical speed of the rock is given, no need to calculate it. Remember: It is thrown horizontally.
     
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