Question

A player passes a 0.600 kg basketball downcourt for a fast break. The ball leaves the player's hands with a speed of 8.30 m/s and slows down to 7.20 m/s at its highest point.
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

a. I assume that the two speeds given are both in the y direction since no angle, θ, given.

We use the equation of motion:

v₁² = v₀² + 2 a (Δx)

Here v₁ is 7.20 m/s, v₀ is 8.30 m/s, and a is gravity (so -g).

(7.2 m/s)² = (8.30 m/s)² - 2 g (Δx)

(8.30 m/s)² - (7.2 m/s)² = 2 g(Δx)

(17.0 m²/s²) / (2g) = Δx

Δx = (17.0 m²/s²) / (2 * 9.8 m/s²)

Δx = 0.87 m

b. Doubling the ball's mass shouldn't affect the maximum height of the ball since:

F_gravity = m * g = m * a

a = g

So the acceleration of the ball would be the same regardless of the mass of the ball.