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Math Help - Range of a projectile

  1. #1
    A riddle wrapped in an enigma
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    Range of a projectile

    Hey guys, I tried to post this on Physicshelpforum, but LaTex is dead.

    Should be a simple solution....but....it's eluding me.


    A batter hits a ball with an initial velocity of v_0 of 100 \text{ f/s } at an angle \theta to the horizontal.


    An outfielder catches the ball 200 \text{ feet } from home plate.


    Find \theta if the range of a projectile is given by the formula:

    R=\frac{1}{32}v_0^2 \sin 2 \theta


    Here's what I've done so far.


    200=\frac{1}{32}(100)^2 \sin 2 \theta


    6400=10000 \sin 2 \theta


    \frac{16}{25}=\sin 2\theta


    \frac{16}{25}=2\sin \theta \cos \theta


    \frac{8}{25}=\sin \theta \cos \theta


    Now, I can't seem to boil this thing down to just one trig function. I know the answer. \theta \approx 20^{\circ}. I just need to know how to get to it.


    Any ideas?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by masters View Post
    Hey guys, I tried to post this on Physicshelpforum, but LaTex is dead.
    Blasphemy!

    Quote Originally Posted by masters View Post
    Should be a simple solution....but....it's eluding me.


    A batter hits a ball with an initial velocity of v_0 of 100 \text{ f/s } at an angle \theta to the horizontal.


    An outfielder catches the ball 200 \text{ feet } from home plate.


    Find \theta if the range of a projectile is given by the formula:

    R=\frac{1}{32}v_0^2 \sin 2 \theta


    Here's what I've done so far.


    200=\frac{1}{32}(100)^2 \sin 2 \theta


    6400=10000 \sin 2 \theta


    \frac{16}{25}=\sin 2\theta
    Let's start from here. Take the inverse sine of both sides:
    \displaystyle sin^{-1} \left ( \frac{16}{25} \right )

    \displaystyle \theta = \frac{1}{2} sin^{-1} \left ( \frac{16}{25} \right ) = 2 \theta

    Taking the inverse sine of 16/25 only gives you a reference angle. So
    \displaystyle \frac{1}{2}(180 - (\text{reference angle})) \approx 70^o

    is also a solution.

    -Dan
    Last edited by topsquark; March 15th 2011 at 01:24 PM.
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  3. #3
    A riddle wrapped in an enigma
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    Well, Dan. It makes perfect sense to me now.

    I was too eager to convert that \sin 2\theta, thus muddying the waters a bit. Thanks for enlightening me.
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