Answer to Question #282854 in Classical Mechanics for OMAR

Question #282854

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?


1
Expert's answer
2021-12-27T08:00:49-0500

α=48°\alpha = 48\degree

h0=2mh_0 = 2m

h1=3.05mh_1 =3.05m

g=9.8ms2g = 9.8\frac{m}{s^2}

s1=7.5s_1= 7.5

v?\vec v-?

s=vxt=vcosαts = v_x*t= v*\cos \alpha* t

t=svcosαt= \frac{s} { v*\cos \alpha}

h1=h0+vytgt22h_1 = h_0+v_yt-\frac{gt^2}{2}

vy=vsinαv_y = v*\sin \alpha

h1=h0+stanαgs22v2cos2αh_1 = h_0+s*\tan \alpha-\frac{gs^2}{2*v^2*\cos^2\alpha}


3.05=2+7.51.119.87.522v20.6723.05 = 2 +7.5*1.11-\frac{9.8*7.5^2}{2*v^2*0.67^2}

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


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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment