Pitching, Catching, and Hitting a Baseball

The impulse-momentum theorem states that the impulse $J$ on an object equals its change in momentum $\Delta p$:

$$J = \Delta p.$$

The impulse is the product of the average force and the time interval:

$$J = F_{\mathrm{pitch}} \cdot \Delta t_{\mathrm{pitch}}.$$

The momentum $p$ of the baseball is the product of its mass and velocity. Since the ball starts from rest, its initial velocity is $0$, so the change in momentum is

$$\Delta p = m_{\mathrm{ball}} \cdot v_{\mathrm{pitch}} - m_{\mathrm{ball}} \cdot 0 = m_{\mathrm{ball}} \cdot v_{\mathrm{pitch}}.$$

Equating impulse and change in momentum gives

$$F_{\mathrm{pitch}} \cdot \Delta t_{\mathrm{pitch}} = m_{\mathrm{ball}} \cdot v_{\mathrm{pitch}}.$$

Solving for the time interval $\Delta t_{\mathrm{pitch}}$:

$$\Delta t_{\mathrm{pitch}} = \frac{m_{\mathrm{ball}} \cdot v_{\mathrm{pitch}}}{F_{\mathrm{pitch}}}.$$

Substituting the given numeric values:

$$\Delta t_{\mathrm{pitch}} = \frac{0.147 \, \mathrm{kg} \cdot 35.7 \, \mathrm{m/s}}{207 \, \mathrm{N}}.$$

Using $1 \, \mathrm{N} = 1 \, \mathrm{kg \cdot m/s^2}$, the units simplify to seconds, and the numerical evaluation gives

$$\Delta t_{\mathrm{pitch}} = 0.025 \, \mathrm{s}.$$