Question

A major league baseball player hits a pitch for a home

run. It was moving at 40.0 m/s toward the batter but
moves at 51.0 m/s away from the batter after being hit. If
the change in velocity happened in 0.34 s, what was the
acceleration of the ball? (Note: Away from the batter is
positive.)

Answer 1

The acceleration of the baseball after being hit is calculated using the change in velocity and the time interval, resulting in an acceleration of approximately 267.65 m/s² away from the batter.

Step-by-step explanation:

Calculating the Acceleration of the Baseball

The problem involves a baseball changing velocity after being hit by the batter. To solve it, we can use the formula for acceleration (a):

a = Δv / Δt

where:

Δt is the change in time

Since away from the batter is considered positive, the change in velocity is:

Δv = vf - vi = 51.0 m/s - (-40.0 m/s) = 91.0 m/s

The time interval during which this change occurs is 0.34 s. So, the acceleration is:

a = 91.0 m/s / 0.34 s ≈ 267.65 m/s²

The acceleration of the baseball is approximately 267.65 m/s² away from the batter.

Answer 2

I got 37.4 sorry if it's not right

Step-by-step explanation:

You set up your triangle so acceleration is at the top then you have m/s on the bottom left then velocity at the bottom right. Then you want to add, subtract, or divide. This is a three step problem.