Question

A baseball travelling horizontally at 41 m/s [S] is hit by a baseball bat, causing its velocity to become 47 m/s [N]. The ball is in contact with the bat for 1.9 ms, and undergoes constant acceleration during this interval. What is that acceleration?

Answer 1

Answer:

$46.3\times 10^3 m/s^2$

Explanation:

We are given that

Initial velocity of baseball=u=-41m/s

Because final and initial velocity are in opposite direction

Final velocity of baseball=v=47m/s

Time=t=1.9 ms=$1.9\times 10^{-3} s$

1 ms=$10^{-3} s$

We have to find the acceleration .

We know that

Acceleration=$a=\frac{v-u}{t}$

Using the formula

Acceleration=$a=\frac{47+41}{1.9\times 10^{-3}}m/s^2$

Acceleration=$46.3\times 10^3 m/s^2$