An Olympic basketball player shoots towards a basket that is 5.54 m horizontally from her and 3.05 m above the floor. The ball leaves her hand 1.52 m above the floor at an angle of 63.0o above the horizontal.
What initial speed should she give the ball so that it reaches the basket and hopefully scores?

Quote:

---

Originally Posted by jenn951

An Olympic basketball player shoots towards a basket that is 5.54 m horizontally from her and 3.05 m above the floor. The ball leaves her hand 1.52 m above the floor at an angle of 63.0o above the horizontal.
What initial speed should she give the ball so that it reaches the basket and hopefully scores?

---

Let velocity = v

Horizontal velocity = v cos 63, vertical initial velocity = v sin 63

Horizontal distance = 5.54, Hence time = t = 5.54/(v cos63)

Vertical distance s = 1.53

Initial verticle velocity = v sin 63, acceleration = a = -g, t = found above

Substitute in the formula: s = ut + 1/2 at^2

1.53 = v sin 63 (5.54/v cos 63) - .5 g (5.54/v cos63)^2

Substitute value of g and solve for v