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Final velocity of a projectile

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball leaves the club at a speed of 14.0 m/s at an angle of 40.0° above the horizontal . It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the peed of the ball just before it lands.

    2. Relevant equations
    ##v_x = v_{0x} = \left \| v_0 \right \|\cos\theta##
    ##v_y^2 = v_{0y}^2 + 2ay##
    ##v = \sqrt{v_x^2 + v_y^2}##
    3. The attempt at a solution

    We begin with ##v = \sqrt{v_x^2 + v_y^2}## since the magnitude of the final velocity is what we desire. Thus, we need to find ##v_x## and ##v_y##. Therefore, we use ##v_x = v_{0x} = \left \| v_0 \right \|\cos\theta## since the velocity in the x-direction does not change from the initial (we're neglecting air resistance). Thus ##v_x = (14 ~m/s)\cos40^{\circ} = 10.7~m/s##. Now,we find ##v_y##, the final velocity in the y-direction. Since time is not metioned, we'll use ##v_y^2 = v_{0y}^2 + 2ay##, specifically ##v_y = -\sqrt{v_{0y}^2 + 2ay}##. Thus ##v_y = -\sqrt{v_{0y}^2 + 2ay} = -\sqrt{(v\sin\theta)^2 + 2(-9.8~m/s^2)(3~m - 0~m)} = -4.71~m/s##. Therefore, the magnitude of the final velocity should be ##v = \sqrt{v_x^2 + v_y^2} = \sqrt{(10.7~m/s)^2 + (-4.71~m/s)^2} = 11.7~m/s ##. Is this the correct answer? If not what am I doing wrong and why?
     
  2. jcsd
  3. Sep 21, 2015 #2

    Doc Al

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    You might want to write this as ##v_y^2 = v_{0y}^2 + 2a\Delta y##

    (You have the change in height being positive.)

    Another way to double check your answer is to use energy principles.
     
  4. Sep 21, 2015 #3

    haruspex

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    Did you mean ##v_{0y}^2 = v_{y}^2 + 2a\Delta y##?
     
  5. Sep 21, 2015 #4

    Doc Al

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    No, I think I had it correct.

    I believe he's using ##v_{0y}## to represent the y-component of the initial velocity.
     
  6. Sep 21, 2015 #5

    haruspex

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    Sorry - my mistake.
     
  7. Sep 21, 2015 #6
    So is the final answer correct or incorrect? I did utilize a change in y since the final position is 3m and the initial is 0m; the change in that is just 3m. Is my approach correct or am I missing something?
     
  8. Sep 21, 2015 #7

    Doc Al

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    Oops... I just realized that I misread the problem. The green was 3m higher, not the tee. So your approach looks fine to me. I'll check the answer in a minute.
     
  9. Sep 21, 2015 #8

    Doc Al

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    Yes, your answer is correct.
     
  10. Sep 21, 2015 #9
    Awesome! Thanks.
     
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