Question
A basketball player shoots a free throw at a 50 degree angle. Assume the ball is released at a height of 1.8 meters, the hoop is 3 meters off of the floor and 4.6 meters away from the shooter. With what velocity should the player release the ball to hit nothing but net?
Answers
let speed of release = x
horizontal speed =u = x cos 50
initial vertical speed Vi = x sin 50
4.6 = (x cos 50) t so t = 4.6/(x cos 50)
3 = 1.8 + (x sin 50) t - 4.9 t^2
1.2 = (x sin 50)(4.6/x cos 50) - 4.9 (4.6^2)/(x^2 cos^2 50)
1.2 = 4.6 tan 50 - 4.9 (4.6^2)/(x^2 cos^2 50)
horizontal speed =u = x cos 50
initial vertical speed Vi = x sin 50
4.6 = (x cos 50) t so t = 4.6/(x cos 50)
3 = 1.8 + (x sin 50) t - 4.9 t^2
1.2 = (x sin 50)(4.6/x cos 50) - 4.9 (4.6^2)/(x^2 cos^2 50)
1.2 = 4.6 tan 50 - 4.9 (4.6^2)/(x^2 cos^2 50)
10.58 m/s