Question

A baseball with a mass of m = 0.145 kg moves with a initial velocity of 40 m/s towards a batter on home plate. The batter swings a bat of mass M = 1 kg and length L = 1 m through an angle of 90 degrees in 0.04s, stricking the ball head-on near the then end of the bat.
a) What is the relative velocity of the end of the bat with respect to the baseball the instant before the two collide? Treat the angular velocity of the bat as constant during the swing.
b) Experts estimate that the furtherse a baseball can be struck by a human is roughly 145 m. Assume the velocity of the struck ball makes and angle of 45 degrees above the horizontal and starts 1.5m above the ground. What is the speed of the ball just after it is struck?

Answer 1

a) Calculate the angular velocity of the bat:

$$\omega=\frac{\phi}{t}=\frac{\pi}{2t}.$$

The linear velocity of the bat becomes

$$v_{bat}=\omega L=\frac{\pi L}{2t}.$$

When the baseball moves toward the end of the bat, the relative velocity of the end of the bat looks like (from the ball's viewpoint):

$$v_{bat.rel.ball}=v_{bat}+v_{ball}=\frac{\pi L}{2t}+v_{ball}=$$

$$=\frac{3.14\cdot1}{2\cdot0.04}+40=79.3\text{ m/s}.$$

b) Now consider an another situation. The ball will go upward with speed

$$v_y=v\text{sin}45^\circ,$$

then it will reach its highest point at

$$h_{max}=\frac{v_y^2}{2g}=\frac{(v\text{sin}45^\circ)^2}{2g}.$$

And this will take

$$t_1=\frac{v\text{sin}45^\circ}{g}.$$

After that, since the ball is struck at height of 1.5 m, falling from the maximum height to reach the ground, the ball must go additional 1.5 meters of height:

$$H=h_{max}+1.5.$$

This will take time of

$$t_2=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\Big[\frac{(v\text{sin}45^\circ)^2}{2g}+1.5\Big]}{g}},$$

i.e. the total time up and down will take

$$t=t_1+t_2.$$

Meanwhile the same time it will take for the ball to cover 145 of horizontal distance with the horizontal component of its speed (if we neglect the drag force):

$$t=\frac{145}{v\text{cos}45^\circ}.$$

Equate time $t$ in the last two equations and solve for $v$:

$$\frac{145}{v\text{cos}45^\circ}=\frac{v\text{sin}45^\circ}{g}+\sqrt{\frac{2\Big[\frac{(v\text{sin}45^\circ)^2}{2g}+1.5\Big]}{g}},$$

<<quite long time required to solve it analytically, so solve it numerically>>

$$v=37.5\text{ m/s}.$$