A basketball player 1.93 m tall wants to make a basket from a distance of 14.3 m

Question

The hoop is at a height of 3.05 m. If he shoots the ball (from 1.93 m) at a 48.7° angle, at what initial speed must he shoot the ball so it goes in the basket without touching the backboard. Neglect air resistance, the acceleration of gravity is 9.8. Answer is m/s

Answer 1

Let origin is hoop. Then motion of ball discrabed by equations: x = 14.3 - v₀cos48.7°t; y = - 1.12 + v₀sin48.7°t -

4.9t²; x = 0; t = 21.667/v₀; y = 0; - 1.12 + 0.751tv₀ - 4.9t² = 0; - 1.12 + 16.28 - 4.9(21.667/v₀)²;

(21.667/v₀)² = 3.0939; v₀ = 21.667/√3.0939 = 12.32 m/s