A basketball player shot a ball. It left his hand at a height of 2.05 m above the court at an angle of 40 degrees relative to horizontal, what initial speed did the ball have if it hit "nothing but net" at the basket 4.50 m away? Assume the rim is 3.05 m above the court.

Question

A basketball player shot a ball.  It left his hand at a height of 2.05 m above the court at an angle of 40 degrees relative to horizontal, what initial speed did the ball have if it hit "nothing but net" at the basket 4.50 m away?  Assume the rim is 3.05 m above the court.

Henry · Answer

Dx = Vo^2*sin(2A)/g = 4.50 m. Vo^2*sin(80)/9.8 = 4.50 0.1Vo^2 = 4.50 Vo^2 = 45 Vo = 6.70 m/s[40o]. = Initial velocity.