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?

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