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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.
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!
$\displaystyle t^2=\frac{6.02\tan25^o-1}{4.9}$Quote:
Originally Posted by Soroban
Oh heck, 1's are trivial in Physics anyway. :)Quote:
Originally Posted by ThePerfectHacker
-Dan
