A ball is hit by a baseball bat straight up at 24.0 m/s. The bat is 1.2 m above the ground How long will it take for the ball to reach its highest point? How high will the baseball travel?

a. 2.45 s b. 30.6 m

Explanation:

a. How long will it take for the ball to reach its highest point?

Using v = u - gt, we find the time it takes the ball to reach the highest point where v = velocity of ball at highest point = 0 m/s (since it momentarily stops), u = initial vertical velocity ball = 24.0 m/s (since it was hit straight up), g = acceleration due to gravity = 9.8 m/s² and t = time taken to reach highest point.

v = u - gt

t = (u - v)/g

t = (24.0 m/s - 0)/9.8 m/s²

t = 24.0 m/s ÷ 9.8 m/s²

t = 2.45 s

b. How high will the baseball travel?

Using y - y₀ = ut - 1/2gt² where y = maximum height of baseball, y₀ = height of bat above the ground where the ball was hit = 1.2 m, u = initial  vertical velocity of ball = 24.0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken to reach highest point = 2.45 s

Substituting these values into the equation, we have

y - y₀ = ut - 1/2gt²

y - 1.2 m = 24.0 m/s × 2.45 s - 1/2 × 9.8 m/s² × (2.45 s)²

y - 1.2 = 58.8 - 29.41

y - 1.2 = 29.39

y = 29.39 + 1.2

y = 30.59 m

y ≅ 30.6 m

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