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Math Help - Basketball player, ballistics

  1. #1
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    Basketball player

    A 2.05 m tall basketball player takes a shot when he is 6.02 m from the basket (at the three-point line). If the launch angle is 25o and the ball was launched at the level of the player's head, what must be the release speed of the ball for the player to make the shot? The basket is 3.05 m above the floor.
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  2. #2
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    Hello, Celia!

    A 2.05 m tall basketball player takes a shot when he is 6.02 m from the basket.
    If the launch angle is 25 and the ball was launched at the level of the player's head,
    what must be the release speed of the ball for the player to make the shot?
    The basket is 3.05 m above the floor.
    Code:
                     *
                 *        *
              *               *
            * 25             | 1
          * - - - - - - - - - *
          |        6.02       |
     2.05 |                   | 2.05
          |                   |
          * - - - - - - - - - *
                   6.02

    We can ignore the bottom half of the diagram.
    The ball must travel 6.02 m horizontally and 1 m vertically (up).

    The equation for projectile motion are:
    . . . v\cos25^o)t" alt="x\:=\v\cos25^o)t" />
    . . . v\sin25^o)t - 4.9t^2" alt="y \:= \v\sin25^o)t - 4.9t^2" />
    where v is the initial velocity (release speed).

    We have: . v\cos25^o)t \:= \:6.02 \\ y \:= \v\sin25^o)t - 4.9t^2\:=\:1\end{array}" alt="\begin{array}{cc}x \:= \v\cos25^o)t \:= \:6.02 \\ y \:= \v\sin25^o)t - 4.9t^2\:=\:1\end{array}" />

    The equations are: . \begin{array}{cc}(1)\;(v\cos25^o)t\:=\:6.02  \\  (2)\;(v\sin25^o)t \:=\:4.9t^2 + 1\end{array}

    Divide (2) by (1): . \frac{(v\sin25^o)t}{(v\cos25^o)t} \:= \:\frac{4.9t^2 + 1}{6.02}\quad\Rightarrow\quad \tan25^o  =\:\frac{4.9t^2 + 1}{6.02}

    Then: . 4.9t^2 + 1 \:=\:6.02\tan25^o\quad\Rightarrow\quad t^2 = \frac{6.02\tan25^o - 1}{4.9} =0.368810633

    Hence: . t\:\approx\:0.6073 seconds.

    From (1), we have: . v\:=\:\frac{6.02}{t\cos25^o}\:=\:\frac{6.02}{0.607  3\cos25^o} \:\approx\:10.94 m/sec



    Edit: Corrected my typo . . . Thanks for the heads-up, TPHacker!
    Last edited by Soroban; June 27th 2006 at 07:11 AM.
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  3. #3
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    Quote Originally Posted by Soroban

    Then: . 4.9t^2 + 1 \:=\:6.02\tan25^o\quad\Rightarrow\quad t^2 = \frac{6.02\tan25^o}{4.9} =0.368810633
    t^2=\frac{6.02\tan25^o-1}{4.9}
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker
    t^2=\frac{6.02\tan25^o-1}{4.9}
    Oh heck, 1's are trivial in Physics anyway.

    -Dan
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