# Physics

• October 8th 2008, 03:31 PM
iz1hp
Physics
An Olympic basketball player shoots towards a basket that is 5.28 m horizontally from her and 3.05 m above the floor. The ball leaves her hand 1.57 m above the floor at an angle of 61.0o above the horizontal.
What initial speed should she give the ball so that it reaches the basket and hopefully scores?

I have no clue how to even start this problem
I know a = -9.81 m/s^2
and Xo = 0
But thats all i know
Thanks
• October 8th 2008, 04:06 PM
skeeter
$\Delta x = v_x \cdot t$

solve for t ...

$t = \frac{\Delta x}{v_x}$

$\Delta y = v_{y0} t - \frac{1}{2}gt^2$

sub in the expression for t ...

$\Delta y = v_{y0} \cdot \frac{\Delta x}{v_x} - \frac{1}{2}g\left(\frac{\Delta x}{v_x}\right)^2$

clean up ...

$\Delta y = v\sin{\theta} \cdot \frac{\Delta x}{v\cos{\theta}} - \frac{g}{2}\left(\frac{\Delta x}{v\cos{\theta}}\right)^2$

$\Delta y = \Delta x \cdot \tan{\theta} - \frac{(\Delta x)^2 g}{2v^2\cos^2{\theta}}$

$v = \sqrt{\frac{(\Delta x)^2 g}{(\Delta x \tan{\theta} - \Delta y)2\cos^2{\theta}}}$
• October 8th 2008, 04:16 PM
iz1hp
$

v_x
$

$

\Delta x
$

how do i get these value?
• October 8th 2008, 04:51 PM
skeeter
$v_x$ is the x-component of your velocity ... $v_x = v\cos{\theta}$

you were given $\Delta x$ in the problem statement.