Physics Question Help?

Question

A 1.80m tall basketball player attempts to make a shot 13.40m from the basket ( the basket is 3.05m high). The accleration of gravity is 9.81 m/s2. If he shoots the ball at a 48 degree angle, what intial speed must he throw the basketball so that it goes through the hoop without striking the backboard. Answer in m/s.

Answer 1

Let V be the velocity in question and Vh and Vv be its horizontal and vertical components respectively.
V= Vh/cos(48) and
V= Vv/sin(48)


Let total time be
t3=t2+2t1
t1- time to reach the basket level from the players height
t2- time in flight from one side to the other at the baskets level.

t2=sqrt(2(H-h)/g)
H -height of the basket
h - height of the player

also the total time
t3= S/Vh
S - horizontal distance from the player to the center of the basket

Now lets put the equation together

t3= S/Vh = t2 + 2t1 in terms of V=Vh/cos (A) and a little algebra

S cos(A)= t2 V + 2 t1 V ; t1=Vv/g = Vsin(A) /g so

S cos(A)= t2 V + 2 (Vsin(A) /g )V we have a quadratic equation

2 (sin(A)/g) V^2 + t2 V - S cos(A)=0

Now just solve for V keeping in mind that t2=sqrt(2(H-h)/g)