A softball having a mass of 0.25 kg is pitched horizontally at 120 km/h. By the time it reaches the plate, it may have slowed by 10%. Neglecting gravity, estimate the average force of air resistance during a pitch. The distance between the plate and the pitcher is about 15 m.

According to the work-energy principle, the net work done by an object is equal to the change in kinetic energy of that object, i.e.,

\(\begin{aligned}{c}\Delta KE = {W_{{\rm{net}}}}\\K{E_{\rm{f}}} - K{E_{\rm{i}}} = {W_{{\rm{net}}}}\end{aligned}\)

Mass of softball is m = 0.25 kg.

Displacement of the softball is s = 15 m.

The initial speed of the softball is as follows:

\(\begin{aligned}{c}{v_{\rm{i}}} = 120\;{\rm{km/h}}\\ = \left( {120\;{\rm{km/h}}} \right) \times \left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right) \times \left( {\frac{{1\;{\rm{h}}}}{{3600\;s}}} \right)\\ = 33.33\;{\rm{m/s}}\end{aligned}\)

The ball has slowed down by 10%. So, the final speed of the softball is as follows:

\(\begin{aligned}{c}{v_{\rm{f}}} = 120\;{\rm{km/h}} - \left( {\frac{{10}}{{100}} \times 120\;{\rm{km/h}}} \right)\\ = 108\;{\rm{km/h}}\\ = \left( {108\;{\rm{km/h}}} \right) \times \left( {\frac{{1000\;{\rm{m}}}}{{1\;{\rm{km}}}}} \right) \times \left( {\frac{{1\;{\rm{h}}}}{{3600\;s}}} \right)\\ = 30\;{\rm{m/s}}\end{aligned}\)

Let F be the average force of air resistance during the pitch. Since air resistance slows down the ball, the force due to air resistance is applied in the direction opposite to the motion of the ball. Therefore, the angle between displacement vector and force vector is \(180^\circ \). So, work done on the softball due to air resistance is as follows:

\(\begin{aligned}{c}W = Fx\cos 180^\circ \\ = - Fx\end{aligned}\) ... (i)

According to the work-energy principle, change in kinetic energy of the softball is equal to the work done by the air resistance on the softball, i.e.,

\(\begin{aligned}{c}\Delta KE = W\\\frac{1}{2}mv_{\rm{f}}^{\rm{2}} - \frac{1}{2}mv_{\rm{f}}^{\rm{2}} = W\\\frac{1}{2}m(v_{\rm{f}}^{\rm{2}} - v_{\rm{i}}^{\rm{2}}) = W\end{aligned}\) ... (ii)

From equations (i) and (ii),

\(\begin{aligned}{c}\frac{1}{2}m(v_{\rm{f}}^{\rm{2}} - v_{\rm{i}}^{\rm{2}}) = - Fx\\F = - \frac{1}{{2x}}m(v_{\rm{f}}^{\rm{2}} - v_{\rm{i}}^{\rm{2}})\end{aligned}\)

On substituting given values in the above equation,

\(\begin{aligned}{c}F = - \frac{1}{{2\left( {15\;{\rm{m}}} \right)}}\left( {0.25\;{\rm{kg}}} \right)\left( {{{(30\;{\rm{m/s}})}^2} - {{\left( {33.3\;{\rm{m/s}}} \right)}^2}} \right)\\ = - \frac{1}{{2\left( {15\;{\rm{m}}} \right)}}\left( {0.25\;{\rm{kg}}} \right)\left( {900 - 1108.9\;{{\rm{m}}^2}{\rm{/}}{{\rm{s}}^2}} \right)\\ = 1.7\;{\rm{N}}\end{aligned}\)

Thus, the average force of air resistance during a pitch is 1.7 N.