Stephanie serves a volleyball from a height of 0.78 m and gives it a initial velocity of +6.6 m/s straight up.

(1): How high will the volleyball go? The acceleration of gravity is 9.81 m/s^2.

(2): How long will it take the ball to reach its maximum height? Answer in units of s

2.2

To solve this problem, we can use the kinematic equations of motion.

(1) The first step is to find the maximum height reached by the volleyball. We need to use the following equation:

vf^2 = vi^2 + 2ad

Where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the displacement.

In this case, the volleyball is thrown straight up, so the final velocity at the maximum height will be 0 m/s. The initial velocity is +6.6 m/s, and the acceleration due to gravity is -9.81 m/s^2 (negative because it acts in the opposite direction to the initial velocity). We want to find the displacement, which is the maximum height.

Plugging in the values into the equation, we have:

0^2 = (6.6)^2 + 2*(-9.81)*d

This simplifies to:

0 = 43.56 - 19.62d

Rearranging the equation, we get:

19.62d = 43.56

Dividing both sides by 19.62, we get:

d = 43.56 / 19.62 ≈ 2.22 m

Therefore, the volleyball will reach a height of approximately 2.22 m.

(2) To find out how long it will take for the ball to reach its maximum height, we can use another equation:

vf = vi + at

Where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.

At its maximum height, the final velocity is 0 m/s. The initial velocity is +6.6 m/s, and the acceleration due to gravity is -9.81 m/s^2. We want to find the time it takes to reach the maximum height.

Plugging in the values, we have:

0 = 6.6 + (-9.81)*t

Rearranging the equation, we get:

9.81t = 6.6

Dividing both sides by 9.81, we get:

t = 6.6 / 9.81 ≈ 0.67 s

Therefore, it will take approximately 0.67 seconds for the ball to reach its maximum height.