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MissTica
2 years ago
12

A 0.145-kg baseball pitched horizontally at 27.0 m/s strikes a bat and pops straight up to a height of 31.5 m. If the contact ti

me between bat and ball is 2.5 ms, calculate the average force between the ball and bat during contact.
Physics
1 answer:
Digiron [165]2 years ago
7 0

Answer:

The average force between the ball and bat during contact is 3006.72  N.

Explanation:

Given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the ball, u = 27 m/s

Time of contact between bat and ball is 2.5 ms, t=2.5\times 10^{-3}\ s

After striking the bat, it pops straight up to a height of 31.5 m. The final velocity of the ball is given by using third equation of motion as :

v^2-u^2=2as

a = -g

And initially, u = 0

v=\sqrt{2gs}

v=\sqrt{2\times 9.8\times 31.5}

v = -24.84 m/s (as it pops straight up)

Let F is the average force between the ball and bat during contact. It is given by :

F=\dfrac{m(v-u)}{t}

F=\dfrac{0.145\times (24.84 -(-27))}{2.5\times 10^{-3}}

F = 3006.72  N

So, the average force between the ball and bat during contact is 3006.72  N. Hence, this is the required solution.                                  

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Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

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