Question

A 2.00 METER TALL BASKETBALL PLAYER ATTEMPTS A GOAL 10.00 METERS FROM THE BASKET(3.05 METERS HIGH). IF HE SHOOTS THE BALL AT A 45.0 DEGREE ANGLE, AT WHAT INITIAL SPEED MUST HE THROW THE BALL SO THAT IS GOES THROUGH THE HOOP WITHOUT STRIKING THE BACKBOARD?

Initial hi=2.0
final hf=3.05
horizontal distance=10

Here are the equations:
hf=Hi + Vsin45*t - 4.9 t^2
and
10= Vcos45*t
In the second equation, solve for t in terms of the numbers and V. Then plug that into the first equation for t, and solve for V. It will be a quadratic, use the quadratic equation.

Answer 1

hf=Hi + Vsin45*t - 4.9 t^2
(v sin 45)t = 1.05 + 4.905 t^2 ___ (1)

(v cos 45)t = 10 ___ (2)

(2)/(1):
10 tan 45 = 1.05 + 4.905 t^2
Solve for t, t = 1.35 s

Substitute t in (2),
(v cos 45)1.35 = 10
v = 10.47 m s^-1