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Homework Help: Vector Question on a golf ball

  1. Oct 3, 2007 #1
    1. The problem statement, all variables and given/known data
    After a golf ball is hit it takes off with an initial speed of 19.1 m/s and at an angle of 42.0° with respect to the horizontal. The golf field is flat and horizontal. Neglecting air resistance how far will the golf ball fly?

    2. Relevant equations
    How high will the golf ball rise?
    How much time will the ball spend in the air?
    How far would the ball fly if the initial speed was doubled?
    How much time would the ball spend in the air in this second case?

    3. The attempt at a solution
    I broke up the Original Vector into their x and y components.

    x component = 14.19
    y component = 12.7

    I know the acceleration is -9.8m/s. but I don't know which equation to use because you're missing the time.
  2. jcsd
  3. Oct 3, 2007 #2


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    A hint for the first question. What would the vertical velocity be when the ball has reached its maximum height?
  4. Oct 3, 2007 #3
    It would be zero no?
  5. Oct 3, 2007 #4


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    Yes it would. Thus, you can use an equation which relates initial velocity, final velocity, acceleration and displacement to one another.

    Edit: Oh, I didn't see the question "how far will the ball fly." If I were answering this, I'd do the "how long will the ball stay in the air" part before the "how far will the ball fly" part-- but that's just me!
    Last edited: Oct 3, 2007
  6. Oct 3, 2007 #5
    Correct. Find the time it takes for the ball to reach its maximum height, using your velocity equation, then plug that time into your distance equation.
  7. Oct 3, 2007 #6
    Alright so I do

    Vf^2 = Vi^2 + 2a x displacement

    Vf = 0
    Vi = 19.1
    a = 19.8

    But I get 18.57 and the online assignment says I'm incorrect.
  8. Oct 3, 2007 #7


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    That's because you used 19.1 for your initial velocity. Remember we are considering the vertical direction, so you need to use the vertical initial velocity. Also, a=-9.8
  9. Oct 3, 2007 #8
    Aha! Alright then, got that answer right. Thanks a bunch.
  10. Oct 3, 2007 #9
    And now I'm stuck once again. I did what l46kok did and found the time using the maximum height and I found that to be 1.29s

    1.29s =t

    But I plug that into x = xo + vot + 1/2a(t)^2
    and I get 32.79 which is wrong. Im not sure if I have the right time..
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