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a pitcher throws a 0.150 - kg baseball, and it approaches the bat at a speed of 44.1 m/s. the bat does 76.5 j of work on the ball in hitting it. ignoring air resistance, determine the speed of the ball after the ball leaves the bat and is 30.2 m above the point of impact.
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Explanation:

Step1: Apply the work - energy theorem

The work - energy theorem states that \(W=\Delta K+\Delta U\), where \(W\) is the work done on the object, \(\Delta K = K_{f}-K_{i}=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}\) is the change in kinetic energy, and \(\Delta U = mgh\) is the change in gravitational potential energy.

We are given \(m = 0.150\space kg\), \(v_{i}=44.1\space m/s\), \(W = 76.5\space J\), \(h = 30.2\space m\), and \(g = 9.8\space m/s^{2}\).

Substitute the expressions for \(\Delta K\) and \(\Delta U\) into the work - energy theorem:

\(W=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}+mgh\)

Step2: Solve for \(v_{f}\)

First, rearrange the equation for \(v_{f}\):

\(\frac{1}{2}mv_{f}^{2}=W+\frac{1}{2}mv_{i}^{2}-mgh\)

Multiply both sides by \(2\) to get:

\(mv_{f}^{2}=2W + mv_{i}^{2}-2mgh\)

Then \(v_{f}^{2}=\frac{2W}{m}+v_{i}^{2}-2gh\)

Substitute the given values:

\(\frac{2\times76.5}{0.150}+(44.1)^{2}-2\times9.8\times30.2\)

\(\frac{153}{0.150}+1944.81 - 591.92\)

\(1020+1944.81 - 591.92\)

\(v_{f}^{2}=2372.89\)

Take the square root of both sides: \(v_{f}=\sqrt{2372.89}\approx48.7\space m/s\)

Answer:

\(48.7\space m/s\)