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Golf ball velocity physics

  1. Jul 5, 2008 #1
    Can someone please help me out! I'm having a lot of trouble getting started in my physics course, and it would be great if I could have the steps of a questions laid out for me. It would make my life so my easier!
    1. The problem statement, all variables and given/known data
    A golfer gives a golf ball a velocity of 48 m/s at an angle of 45° with the
    horizontal.
    (a) what is the vertical component of the ball's initial velocity?
    (b) How long is the ball in the air?
    (c) What is the horizontal distance covered by the ball while in flight?

    2. Relevant equations




    3. The attempt at a solution
     
    Last edited by a moderator: May 11, 2014
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  3. Jul 5, 2008 #2

    nicksauce

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    Re: please!!

    Ok well let's start with part a. How would you approach the problem?
     
  4. Jul 5, 2008 #3
    Re: please!!

    sin of 45? 33.9 m/s? I really have no idea.
     
  5. Jul 5, 2008 #4

    nicksauce

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    Re: please!!

    Yes that's correct, but it helps if you understand why. You should draw an ABC triangle, with A being 45 degrees, B being 45 degrees and C being 90 degrees, where AB = 48m/s, AC = the x component of the velocity and BC = the y component of the velocity. Look at this and convince yourself that the x component must be 48cos(45) and the y component must be 48sin(45). Also convince yourself that this is a fair way to represent vectors. If you can't do so, read your class notes or textbook or whatever.

    Now for part b. You should be familiar with the equations listed here http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html. If you are not, once again, read your notes/textbook. Now tell me, what are your known variables, what is your unkown variable(ie what you are trying to solve for), and which equation is appropriate to calculate the unkown variable.
     
  6. Jul 5, 2008 #5
    Re: please!!

    initial v= 48 m/s
    a= 9.80 m/s2

    You could use d=vt + 1/2at2 (squared)? But I don't have the distance travelled.
     
  7. Jul 5, 2008 #6

    nicksauce

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    Well first of all, you should only consider the y components. Thus Initial v = 33.9m/s. Now because "How long is the ball in the air" implies that the ball is now the ground, you know final y = 0. Use that equation and you'll get two solutions. Omit the one that is t=0.
     
  8. Jul 5, 2008 #7
    Re: please!!

    Could I not use v = vinitial + at ?
    Would the horizontal distance travelled by 118 m?
     
  9. Jul 5, 2008 #8

    nicksauce

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    I imagine you could. However it would be difficult since you don't know what v final is. You are given that y final is 0, and it is always a good idea to use your givens.

    Maybe. How did you get that?
     
  10. Jul 5, 2008 #9
    Re: please!!

    I used v = vinitial + at and got 3.50 s
    then I multipled 33.9 m/s by 3.50 s to get 118 m
    does that work?
     
  11. Jul 5, 2008 #10

    nicksauce

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    I have no idea how using v = vinitial + at you got that the ball was in the air for 3.50 seconds. For one, what do you use for v final?
     
  12. Jul 5, 2008 #11
    Re: please!!

    0? But I guess that doesn't work.
     
  13. Jul 5, 2008 #12

    nicksauce

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    No it doesn't. Velocity and position should be treated as continous variables in kinematics problems. What this means is that you can't suddenly have V go from a non-zero number to zero without taking on all intermediate values. Imagine you drop a ball, and it is within a foot of the ground, its velocity have some value V. Now imagine it's within 6 inches of the ground; its velocity will now be greater in magnitude since it is accelerating. Now imagine it's within 1 inch of the ground; the magnitude of its velocity will be even greater still. The value of v_final can be thought of the value that these velocities approaches as the distances gets closer and closer to the ground.

    Now as I said before, you don't know v_final. But you do know y_final. Use it, and the appropriate formula to find t.
     
    Last edited: Jul 5, 2008
  14. Jul 5, 2008 #13
    Re: please!!

    Okay, great. Thanks so much!! I have one other question, if you don't mind..
    A girl throws a rock horizontally from the top of a cliff 98 m high, with a horontal velocity of 27 m/s.
    a) how many seconds was the rock in the air?
    b) how far out from the base of the cliff does the rock land?

    for a, would I just divide the height of the cliff by the velocity for a time fo 3.63 s?
     
  15. Jul 5, 2008 #14

    Integral

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    Re: please!!

    This will get you your time of flight if you set it up right. You need to set up your coordinate system correctly, note that the acceleration due to gravity acts in the opposite direction as your initial velocity. So the correct equation would be

    [tex] v = v_{iy} -at [/tex]

    At the peak of motion your y velocity will be 0, use that with the initial y velocity to the time to the top. So what will be your total time?

    Given the time of flight use your constant x velocity to get the distance traveled.
     
  16. Jul 5, 2008 #15

    nicksauce

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    No that will not work. You are interested in the vertical displacement, but the initial velocity is the horizontal direction. What you said would only work if the initial velocity was in the vertical direction, and there was no vertical acceleration.

    What do you know? You know y_initial = 98m, y_final = 0, vy_initial = 0, vx_initial = 27m/s, a_y = -9.8m/s^2, a_x = 0. Can you find an equation into which you put these together to calculate t?
     
  17. Jul 5, 2008 #16
    Re: please!!

    How do I incorporate yinitial and yfinal into an equation?
     
  18. Jul 5, 2008 #17

    nicksauce

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    y_final = y_initial + vy_initial*t + 1/2 * a_y * t^2

    You know all of these except for t.
     
  19. Jul 5, 2008 #18

    Integral

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    Re: please!!

    He has the initial velocity and the angle, therefore he has the initial velocity in the vertical (y). Use that information in the equation I posted.


    Edit: Nick you have a PM
     
  20. Jul 5, 2008 #19

    Integral

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    This is very misleading. The velocity of the ball just before, or when, it hits the ground is the SAME as the initial velocity. For the vertical motion the velocity is 0 at the highest point of motion.

    Nick, please see your PM.
     
  21. Jul 5, 2008 #20

    nicksauce

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    Re: please!!

    I wrote that final y = 0, meaning the final y position is 0. What you just wrote has nothing to do with that. I certainly do not think the velocity when the ball hits the ground is 0. In fact I made a point of arguing against this in a previous post, when the OP had this misconception.
     
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