A 0.145-kg baseball pitched horizontally at 27.0 m/s strikes a bat and pops straight up to a height of 31.5 m. If the contact time between bat and ball is 2.5 ms, calculate the average force between the ball and bat during contact.

According to Newton’s second law, the rate of change of momentum of an object is equal to the net force applied to it.

In this problem, the average force between the ball and the bat is calculated by applying Newton’s second law on the motion of the ball along with horizontal and vertical directions separately.

Mass of the baseball,\(m = 0.145\;{\rm{kg}}\).

The initial velocity of the baseball,\({u_1} = 27.0\;{\rm{m/s}}\).

Vertical height traveled by the ball, h = 31.5 m.

Contact time between bat and ball,\(\Delta t = 2.5\;{\rm{ms}} = 2.5 \times {10^{ - 3}}\;{\rm{s}}\).

The ball moves in the vertical direction on striking the bat. Thus, the velocity of the ball along a horizontal direction at the time of strike with the bat will be\({v_1} = 0\;{\rm{m/s}}\).

During the motion of the ball along the vertical direction, the velocity of the ball at the highest point will be\({v_2} = 0\;{\rm{m/s}}\).

Let the initial velocity of the bullet after striking the bat along the vertical direction is \({u_2}\).

During the vertical motion of the ball, the initial potential energy of the ball at the time of strike with the bat will be zero.

On applying the law of conservation of mechanical energy for the vertical motion of the ball,

\(\begin{array}{c}K{E_1} + P{E_1} = K{E_2} + P{E_2}\\\frac{1}{2}mu_2^2 + mg(0) = \frac{1}{2}mv_2^2 + m{\rm{g}}h\\\frac{1}{2}\left( {0.145\;{\rm{kg}}} \right)u_2^2 + 0 = \frac{1}{2}\left( {0.145\;{\rm{kg}}} \right)\left( {0\;{\rm{m/s}}} \right) + \left( {0.145\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {31.5\;{\rm{m}}} \right)\\\frac{1}{2}\left( {0.145\;{\rm{kg}}} \right)u_2^2 = \left( {0.145\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {31.5\;{\rm{m}}} \right)\\u_2^2 = 2\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {31.5\;{\rm{m}}} \right)\\{u_2} = 24.85\;{\rm{m/s}}\end{array}\)

Thus, the ball moves up straight after striking the bat with a velocity of \(24.85\;{\rm{m/s}}\).

From Newton’s second law, the average force between the ball andbat when the ball is moving along the horizontal direction is as follows:

\(\begin{array}{c}{{\bar F}_1} = \frac{{\Delta {p_1}}}{{\Delta t}}\\ = \frac{{m\left( {{v_1} - {u_1}} \right)}}{{\Delta t}}\\ = \frac{{\left( {0.145\;{\rm{kg}}} \right)\left( {0\;{\rm{m/s}} - 27.0\;{\rm{m/s}}} \right)}}{{\left( {2.5 \times {{10}^{ - 3}}\;{\rm{s}}} \right)}}\\ = - 1566\;{\rm{N}}\end{array}\)

From Newton’s second law, the average force between the ball andbat when the ball is moving along the horizontal direction is as follows:

\(\begin{array}{c}{{\bar F}_2} = \frac{{\Delta {p_2}}}{{\Delta t}}\\ = \frac{{m\left( {{v_2} - {u_2}} \right)}}{{\Delta t}}\\ = \frac{{\left( {0.145\;{\rm{kg}}} \right)\left( {0\;{\rm{m/s}} - 24.85\;{\rm{m/s}}} \right)}}{{\left( {2.5 \times {{10}^{ - 3}}\;{\rm{s}}} \right)}}\\ = - 1441.3\;{\rm{N}}\\ = - 1441\;{\rm{N}}\end{array}\)

The magnitude of the average force between the ball and the bat during contact is as follows:

\(\begin{array}{c}\bar F = \sqrt {\bar F_1^2 + \bar F_2^2} \\ = \sqrt {{{\left( { - 1566\;{\rm{N}}} \right)}^2} + {{\left( { - 1441\;{\rm{N}}} \right)}^2}} \\ = 2128.11\;{\rm{N}}\\ = 2.1\; \times {10^3}\;{\rm{N}}\end{array}\)

If average force makes an angle\(\theta \)with the horizontal, then

\(\begin{array}{c}\tan \theta = \frac{{{{\bar F}_2}}}{{{{\bar F}_1}}}\\ = \frac{{\left( { - 1441\;{\rm{N}}} \right)}}{{\left( { - 1566\;{\rm{N}}} \right)}}\\ = 0.92\\\theta = {\tan ^{ - 1}}\left( {0.92} \right)\\ = 42.6^\circ \\ \approx 43^\circ \end{array}\)

Thus, the average force between the ball and bat during contact is \(2.1\; \times {10^3}\;{\rm{N}}\), which acts in a direction that makes an angle of\(43^\circ \) with the horizontal.