a tennis ball is thrown vertically upward with an initial velocity of +8.0 m/s

a) what is the velocity of the ball when it returns to its starting point?

b) how long is the ball in the air?

c) how high does the ball reach?

To answer these questions, we can use the equations of motion for objects in free fall. First, let's list the given information:

Initial velocity (vi) = +8.0 m/s (upward)
Acceleration (a) = -9.8 m/s^2 (due to gravity, pointing downward)

a) To find the velocity of the ball when it returns to its starting point, we can use the fact that the final velocity (vf) is equal to the negative of the initial velocity because the ball will have reversed its direction. So, vf = -8.0 m/s.

b) To find the time the ball is in the air, we can use the equation: vf = vi + at. Plugging in the given values, we have -8.0 m/s = 8.0 m/s - 9.8 m/s^2 * t. Solving for t, we get:

-8.0 m/s + 8.0 m/s = -9.8 m/s^2 * t
0 = -9.8 m/s^2 * t

Since the initial and final velocities are equal in magnitude but opposite in sign, we can conclude that the ball will return to its starting point in the same amount of time it took to go up. Thus, t = 0.

c) To find the maximum height the ball reaches, we can use the equation: vf^2 = vi^2 + 2aΔy. Since the final velocity at the highest point is 0 m/s, we have:

0^2 = 8.0 m/s^2 + 2 * -9.8 m/s^2 * Δy
0 = 64.0 m^2/s^2 - 19.6 m/s^2 * Δy

Solving for Δy, we get:

19.6 m/s^2 * Δy = 64.0 m^2/s^2
Δy = 64.0 m^2/s^2 / 19.6 m/s^2
Δy ≈ 3.27 m

Therefore, the ball reaches a maximum height of approximately 3.27 meters.