Question

1. A baseball is hit straight up at a speed of 45 meters per second.

a. What is the maximum height of the ball before it falls back down?

b. How long is the baseball in the air?​

Answer 1

a) 103.32 m

b) 9.18 s

Step-by-step explanation:

a) Let's use the knowledge that at the top of its trajectory, the baseball will have a final velocity of 0 m/s.

The acceleration due to gravity is -9.8 m/s², assuming the upwards direction is positive and the downwards direction is negative.

The initial velocity of the baseball is 45 m/s.

We are trying to find the vertical displacement of the baseball, Δx, and we have the variables v, a, and v₀.

Find the constant acceleration equation that contains all four of these variables:

• v² = v₀² + 2aΔx

Substitute the known values into the equation.

• (0)² = (45)² + 2(-9.8)Δx
• 0 = 2025 - 19.6Δx
• -2025 = -19.6Δx
• Δx = 103.32

The maximum height of the ball before it falls back down is 103.32 m.

b) Now we want to solve for time t. Find a constant acceleration equation that contains three known variables.

• v = v₀ + at

Substitute known values into this equation.

• 0 = 45 + (-9.8)t
• -45 = -9.8t
• t = 4.59183673

Remember that this is only half of the baseball's flight since we are using the final velocity for when the ball is halfway through its trajectory.

To solve for the total time the baseball is in the air, double the time t we solved for.

• 2t = 9.18367346

The baseball is in the air for 9.18 s.