• Jun 26th 2006, 07:10 PM
Celia
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.
• Jun 26th 2006, 08:45 PM
Soroban
Hello, Celia!

Quote:

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:
. . . $\displaystyle x\:=\:(v\cos25^o)t$
. . . $\displaystyle y \:= \:(v\sin25^o)t - 4.9t^2$
where $\displaystyle v$ is the initial velocity (release speed).

We have: .$\displaystyle \begin{array}{cc}x \:= \:(v\cos25^o)t \:= \:6.02 \\ y \:= \:(v\sin25^o)t - 4.9t^2\:=\:1\end{array}$

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

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

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

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

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

Edit: Corrected my typo . . . Thanks for the heads-up, TPHacker!
• Jun 27th 2006, 05:16 AM
ThePerfectHacker
Quote:

Originally Posted by Soroban

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

$\displaystyle t^2=\frac{6.02\tan25^o-1}{4.9}$
• Jun 27th 2006, 12:50 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
$\displaystyle t^2=\frac{6.02\tan25^o-1}{4.9}$

Oh heck, 1's are trivial in Physics anyway. :)

-Dan