
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

$\displaystyle \Delta x = v_x \cdot t$
solve for t ...
$\displaystyle t = \frac{\Delta x}{v_x}$
$\displaystyle \Delta y = v_{y0} t  \frac{1}{2}gt^2$
sub in the expression for t ...
$\displaystyle \Delta y = v_{y0} \cdot \frac{\Delta x}{v_x}  \frac{1}{2}g\left(\frac{\Delta x}{v_x}\right)^2$
clean up ...
$\displaystyle \Delta y = v\sin{\theta} \cdot \frac{\Delta x}{v\cos{\theta}}  \frac{g}{2}\left(\frac{\Delta x}{v\cos{\theta}}\right)^2$
$\displaystyle \Delta y = \Delta x \cdot \tan{\theta}  \frac{(\Delta x)^2 g}{2v^2\cos^2{\theta}}$
$\displaystyle v = \sqrt{\frac{(\Delta x)^2 g}{(\Delta x \tan{\theta}  \Delta y)2\cos^2{\theta}}}$

$\displaystyle
v_x
$
$\displaystyle
\Delta x
$
how do i get these value?

$\displaystyle v_x $ is the xcomponent of your velocity ... $\displaystyle v_x = v\cos{\theta}$
you were given $\displaystyle \Delta x$ in the problem statement.