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A pitcher throws a baseball at 31.0 m/s. The batter hits it and gives it a velocity of 40.0 m/s in the opposite direction. If the ball and bat are in contact for 4.00 x 10^-2 s, what is the average acceleration of the ball during this interval?

Respuesta :

Answer:

Acceleration will be 1775 [tex]m/sec^2[/tex]

Explanation:

We have given that pitcher throws the ball with speed of 31 m/sec

And the ball bounce back with a speed of 40 m/sec

Time [tex]t=4\times 10^{-2}sec[/tex]

From first equation of motion we know that v = u+at

So acceleration [tex]a=\frac{v-u}{t}=\frac{40-(-31)}{4\times 10^{-2}}=1775m/sec^2[/tex] (for calculation we take -31 because it is in opposite direction )

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