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Projectile motion problem (on the moon)

  1. Oct 3, 2004 #1
    actually, i have 2 problems i'm having terrible problems with:

    #1: i throw a ball with an intial SPEED of 15m/s at an unknown angle. there's a 2metre high wall 25 metres in front of me. between which 2 angles should i aim the ball for it to fly above this wall?

    #2: i throw a ball at unknown speed at an inclination of 30 degrees, and it goes over a fence (of unknown height....) 80 metres away, and it's total time of flight is 4.72 seconds (also note that the ball does not necessarily end up on the same ground level as where i threw it from. it could have gone under a cliff, or up a hill...) if this exact scenario took place on the moon (which has gravity of 1.627m/s^2) instead of on earth, what would the time of flight be?
     
  2. jcsd
  3. Oct 4, 2004 #2
    #1

    Assume it occurred on the moon.
    x-component:
    s=ut
    [tex]25=(15cos\theta)t[/tex]…(1)
    [tex]t=\frac{5}{3cos\theta}[/tex]
    y-component:
    [tex]s=ut-\frac{g}{2}t^2>2[/tex]…(2)
    (1) into (2) and simplify with g=1.627 meter per second per second:
    [tex]225cos\theta sin\theta-20.3375>18cos^2\theta[/tex]…(3)
    [tex]Let\ cos\theta=x;\ then\ sin\theta=\sqrt{1-x^2}[/tex]and substitute into (3) and simplify :
    [tex]x\sqrt{1-x^2}>0.08x^2+0.09039[/tex]
    Square both sides and simplify :
    [tex]1.0064x^4-0.98554x^2+0.008170<0[/tex]
    Find the roots using quadratic formula:
    [tex]x=\pm 0.9854\ or\ \pm0.09144[/tex]
    Draw a line number and identify the area which satisfy the inequality :
    [tex]0.09144<x<0.9854[/tex]
    [tex]0.09144<cos\theta<0.9854[/tex]
    [tex]cos^{-1}0.9854<\theta<cos^{-1}0.09144[/tex]
    [tex]9.80^0<\theta<84.7^0 [/tex]
     
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