Question

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 leaves 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.

Answer 1

Answer:

$-7.352\times 10^3$N

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=$-7.352\times 10^3$N