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Throwing a ball

  1. Sep 28, 2008 #1
    1. The problem statement, all variables and given/known data

    By throwing a ball at an angle of 35°, a girl can throw it a maximum horizontal distance of R = 42 m on a level field. How far can she throw the same ball vertically upward? Assume that her muscles give the ball the same speed in each case.

    How far? in meters

    2. Relevant equations

    a^2 + b^2 = c ^2 or ..
    y int + (v int * sin beta) t - (1/2) g*t^2 ??

    3. The attempt at a solution

    I drew the first triangle hyp = 24.19, adj = 12.01, opp = 21 with angles of 35, 90 and 55

    Is that the right start of the problem?
     
    Last edited: Sep 28, 2008
  2. jcsd
  3. Sep 28, 2008 #2
    Are you given the initial velocity of the projectile? It appears not from your given problem statement, but you can calculate it with the equation for the range of a projectile:

    [tex] R = \frac{v^{2}_{0}}{g}sin2\theta [/tex]

    Does this help?
     
  4. Sep 28, 2008 #3
    from that equation, I got initial velocity = 18.92

    sin (35) *2 = 1.15
    42/1.15 = 36.52
    36.52 * 9.8 = 357.91
    st root of 357.91 = 18.92

    Is that the correct answer then?
     
    Last edited: Sep 28, 2008
  5. Sep 28, 2008 #4
    No, this equation will yield only the initial velocity, which is what you need to determine how high the projectile can be thrown vertically. Also, you have a mistake.

    When [tex] \theta = 35 [/tex],

    [tex] 2\theta = 70 [/tex]

    So you want sin(70), not sin(35) * 2. Do you see? If not, see that they are different by inputting them into the calculator.
     
  6. Sep 28, 2008 #5
    Thanks for your help buffordboy!
     
  7. Sep 28, 2008 #6
    No problem, did you get the final answer though (maximum height)?
     
  8. Sep 28, 2008 #7
    I think I got it.

    Do I use v_t = at + int velocity to get the time

    0 = -9.8(t) + 20.93; t = 2.13 s

    then I plug that into
    x(t) = x_o + v_o*t + .5at

    x(t) = 42 + 20.93(2.13) + .5 (-9.8)(2.13^2)

    to get the maximum height?

    which I got is 64.35 m, am I correct or is it completely off?
     
    Last edited: Sep 28, 2008
  9. Sep 28, 2008 #8
    Your equation to get the time was the correct step, and your time value t is correct as well.

    Now, the second equation has an error. You substituted 42m into the equation; this value is the range (maximum x-distance) for the given problem.

    You want to know the maximum vertical distance; this is the y-direction. Use

    y(t) = y_o + v_o*t + .5at

    where y_o = 0; this says the initial height of the ball is at the origin, so when you find y(t) with t = 2.13 s you will find the maximum height y_max = 22.35m.

    Note that y_o could be y_o = 42 m; this says that the initial height of the ball is 42 m above the origin. With this y_o you will find that y(t) = 64.35 m for t = 2.13 s. But y_max = y(t) - y_o = 64.35 m - 42 m = 22.35 m. Same answer, even though we used two different points as our origin. Does this make sense?
     
  10. Sep 28, 2008 #9
    yes, thank you very much! how do I mark this thread as solved?
     
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