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:

. . . $\displaystyle x\:=\v\cos25^o)t$

. . . $\displaystyle y \:= \v\sin25^o)t - 4.9t^2$

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!