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?

Respuesta :

Answer:

[tex]46.3\times 10^3 m/s^2[/tex]

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=[tex]1.9\times 10^{-3} s[/tex]

1 ms=[tex]10^{-3} s[/tex]

We have to find the acceleration .

We know that

Acceleration=[tex]a=\frac{v-u}{t}[/tex]

Using the formula

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

Acceleration=[tex]46.3\times 10^3 m/s^2[/tex]

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