Answer to Question #282854 in Classical Mechanics for OMAR

Question

A basketball player practices shooting three-pointers from a distance of 7.50 m from the hoop, releasing the ball at a height of 2.00 m above the floor. A standard basketball hoop’s rim top is 3.05 m above the floor. Th e player shoots the ball at an angle of 48.0° with the horizontal. At what initial speed must she shoot to make the basket?

Answer 1

$\alpha = 48\degree$

$h_0 = 2m$

$h_1 =3.05m$

$g = 9.8\frac{m}{s^2}$

$s_1= 7.5$

$\vec v-?$

$s = v_x*t= v*\cos \alpha* t$

$t= \frac{s} { v*\cos \alpha}$

$h_1 = h_0+v_yt-\frac{gt^2}{2}$

$v_y = v*\sin \alpha$

$h_1 = h_0+s*\tan \alpha-\frac{gs^2}{2*v^2*\cos^2\alpha}$

$3.05 = 2 +7.5*1.11-\frac{9.8*7.5^2}{2*v^2*0.67^2}$

$v = 9.23\frac{m}{s}$

$\text{Answer: }9.23\frac{m}{s}$