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Max. height and range problem I can't crack

  1. Sep 21, 2004 #1
    19.[2pt]
    On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 4.00 times as far as he would have on earth, given the same initial velocities on both planets. The ball is launched at a speed of 42.6m/s at an angle of 28.2o above the horizontal. When the ball lands, it is at the same level as the tee. On the distant planet, what is the maximum height of the ball?
    Answer:

    20.[2pt]
    What is the range of the ball?
    Answer:


    its so hard to do without earth gravity... i can't figure it out!! :-( plz help!
     
  2. jcsd
  3. Sep 21, 2004 #2
    You're right, it is hard to do without a gravitational constant. Can you find the gravitational constant from the info you have?
     
  4. Sep 22, 2004 #3

    Tide

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    The range is: [itex]R = \frac {v_0^2 \sin {2\theta}}{2g}[/itex] so it's easy to infer that g on the planet is one fourth that on Earth.

    The maximum height is [itex]H = \frac {v_0^2 \sin ^2 \theta}{2g}[/itex] and you should have no trouble doing the rest.
     
  5. Sep 22, 2004 #4
    the trig identity is sin2(thet) is 2sin(thet)cos(thet) right?

    my ansewrs not coming out correct.... i quadroople checked the algebra
     
  6. Sep 22, 2004 #5

    HallsofIvy

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    It's difficult for us to comment on what you have done when you haven't shown us what you have done.
     
  7. Sep 22, 2004 #6
    oh, sorry.

    took R = v0^2 * sin2(theta) /2g

    turned it into v0^2 * 2 * sin(theta) * cos(theta) / 2g (trig identity)

    crossed out the 2s

    plugged in 28 for theta, 42.6 for v0, and 2.45 (.25*9.8) for G. - solved.

    got 307.042 meters - got the question wrong...

    i guess it doesnt matter now because i missed the due date for the problem, but id still like to figure out how to do it for future reference.

    im not exactly clear on why g is .25...
     
  8. Sep 22, 2004 #7

    Tide

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    Start with
    [tex]R = \frac {v_0^2}{2g} \sin 2\theta[/tex]
    and with identical initial speed and angles just set up the proportion
    [tex]\frac {R_1}{R_2} = \frac {g_2}{g_1}[/tex]
    Your statement was that [itex]R_2 = 4 R_1[/itex] so [itex]g_2 = \frac {1}{4} g_1[/itex].
     
  9. Sep 22, 2004 #8
    ohhh......... thanks for clearing that up!! :smile:
     
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