A pitcher throws a 0.198-kg baseball, and it approaches the bat at a speed of 57.5 m/s. The bat does 85.1 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 32.7 m above the point of impact.
To determine the speed of the ball after it leaves the bat, we can use conservation of energy.
First, we need to find the potential energy of the ball when it is 32.7 m above the point of impact. The potential energy (PE) can be calculated using the formula:
PE = m * g * h
Where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height above the point of impact.
Given that the mass of the ball (m) is 0.198 kg and the height (h) is 32.7 m, we can calculate:
PE = 0.198 kg * 9.8 m/s² * 32.7 m
= 63.4 J
Next, we can use the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy (KE). In this case, the bat does 85.1 J of work on the ball.
Work (W) = Change in kinetic energy (ΔKE)
So, we can write:
W = ΔKE
85.1 J = KE final - KE initial
Since the ball is initially at rest, its initial kinetic energy (KE initial) is 0. So, we can rewrite the equation as:
85.1 J = KE final - 0
85.1 J = KE final
Thus, the final kinetic energy (KE final) is 85.1 J.
Finally, we can use the formula for kinetic energy to calculate the final velocity (v) of the ball:
KE = 0.5 * m * v^2
Rearranging the equation, we get:
v = √(2 * KE / m)
Substituting the values, we have:
v = √(2 * 85.1 J / 0.198 kg)
= √(430 J/kg)
≈ 20.7 m/s
Therefore, the speed of the ball after it leaves the bat and is 32.7 m above the point of impact is approximately 20.7 m/s.