Question

A basketball leaves a player's hands at a height of 2.20 m above the floor. The basket is 3.05 m above the floor. The player likes to shoot the ball at a 40.0 ∘ angle.
If the shot is made from a horizontal distance of 6.60 m and must be accurate to ±0.23m (horizontally), what is the range of initial speeds allowed to make the basket?

Answer 1

The equation of the trajectory of the ball

y=y_0+x tan⁡〖θ-(gx^2)/(2v_0^2 cos^2⁡θ )〗

In our case

3.05=2.20+(6.60±0.23)tan⁡〖40.0°-(9.81×〖(6.60±0.23)〗^2)/(2v_0^2 cos^2⁡〖40.0°〗 )〗

The solution of this equation

v_0=(8.81±0.13) m/s=(8.68-8.94) m/s