Question

A player passes a 0.600-kg basketball down court for a fast break. The ball leaves the player's hands with a speed of 8.70 m/s and slows down to 7.10 m/s at its highest point. Part A Ignoring air resistance, how high above the release point is the ball when it is at its maximum height?B) How would doubling the ball's mass affect the result in part (a)?

Answer 1

1.29 m

Step-by-step explanation:

mass of ball = 0.6 kg

initial velocity, u = 8.7 m/s

final velocity, v = 7.1 m/s

acceleration due to gravity, g = - 9.8 m/s^2

(a) Let the ball reaches to a height of h.

Use third equation of motion

`v^(2)=u^(2)+2as`

`7.1^(2)=8.7^(2)- 2 * 9.8 * h`

h = 1.29 m

Thus, the maximum height attained by the ball is 1.29 m.