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rodikova [14]
4 years ago
10

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 43.0 m/s; when it leav

es the bat, the ball is traveling to the left at an angle of 31.0 degree above horizontal with a speed of 51.0 m/s. The ball and bat are in contact for 1.71 ms.
Find the horizontal component of the average force on the ball. Take the x-direction to be positive to the right.
Physics
1 answer:
kotykmax [81]4 years ago
6 0

Answer:

-7.352\times 10^3N

Explanation:

We are given that

Mass of baseball=m=0.145 kg

Initial velocity of ball=u=43 m/s

\theta=31^{\circ}

Speed, v=51 m/s

Time, t=1.71 ms=1.71\times 10^{-3} s

1 ms=1.71\times 10^{-3} s

Horizontal component of the average force on the ball=\frac{m(vcos\theta-u)}{t}

Horizontal component of the average force on the ball=\frac{0.145(-51cos31-43)}{1.71\times 10^{-3}}

Horizontal component of the average force on the ball=-7.352\times 10^3N

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