A pitcher throws a 0.168-kg baseball, and it approaches the bat at a speed of 45.2 m/s. The bat does 86.6 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 22.5 m above the point of impact.

To determine the speed of the ball after it leaves the bat and is 22.5m above the point of impact, we can use the principle of conservation of energy.

The total energy of the ball can be broken down into two parts: kinetic energy (KE) and potential energy (PE). Initially, the ball only has kinetic energy, and after it leaves the bat, it has both kinetic and potential energy.

The kinetic energy of the ball before it leaves the bat can be calculated using the equation:

KE = (1/2)mv^2

where m is the mass of the ball (0.168 kg) and v is its initial velocity (45.2 m/s).

KE = (1/2)(0.168 kg)(45.2 m/s)^2
KE = 171.944 J

Now, let's calculate the potential energy of the ball after it reaches a height of 22.5m. The potential energy (PE) can be calculated using the equation:

PE = mgh

where m is the mass of the ball (0.168 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the ball above the point of impact (22.5 m).

PE = (0.168 kg)(9.8 m/s^2)(22.5 m)
PE = 34.314 J

Since energy is conserved, the total energy before and after remains the same. Therefore, the total energy after the ball leaves the bat is:

Total energy after = KE + PE = 171.944 J + 34.314 J = 206.258 J

Now, we need to calculate the final velocity (v) of the ball after it leaves the bat. The total energy can be expressed as the sum of kinetic and potential energy:

Total energy = (1/2)mv^2 + mgh

Rearranging the equation and solving for v gives:

v = sqrt(2g(h + (KE/m)))

v = sqrt(2 * 9.8 m/s^2 * (22.5 m + 171.944 J / (0.168 kg))
v = sqrt(2 * 9.8 * (22.5 + 171.944 / 0.168)) m/s
v = sqrt(9.8 * (22.5 + 1026.7188)) m/s
v = sqrt(9.8 * 1049.2188) m/s
v = sqrt(10242.26544) m/s
v ≈ 101.2 m/s

Therefore, the speed of the ball after it leaves the bat and is 22.5m above the point of impact is approximately 101.2 m/s.

To determine the speed of the ball after it leaves the bat, we need to apply the principle of conservation of mechanical energy. Since the problem tells us to ignore air resistance, the only forms of mechanical energy we need to consider are kinetic energy and gravitational potential energy.

1. First, let's calculate the kinetic energy of the ball when it approaches the bat. The formula for kinetic energy is K = 0.5 * m * v^2, where m is the mass of the ball and v is its velocity. Plugging in the given values, we have:
K1 = 0.5 * 0.168 kg * (45.2 m/s)^2

2. Next, let's calculate the gravitational potential energy of the ball when it is 22.5 m above the point of impact. The formula for gravitational potential energy is U = m * g * h, where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height above the reference point (in this case, the point of impact). Plugging in the given values, we have:
U = 0.168 kg * 9.8 m/s^2 * 22.5 m

3. According to the conservation of mechanical energy, the total mechanical energy of the ball before and after the point of impact remains constant. Therefore, the sum of its kinetic energy and gravitational potential energy is constant. Mathematically, we can write:
K1 + U = K2 + U2,
where K1 is the initial kinetic energy, U is the initial gravitational potential energy, K2 is the final kinetic energy, and U2 is the final gravitational potential energy.

4. Since we are interested in finding the final kinetic energy (which will allow us to determine the speed of the ball), we can rewrite the equation as:
K2 = K1 + U - U2.

5. Given that the bat does 86.6 J of work on the ball, we can equate the work done with the change in kinetic energy. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. Mathematically, we can write:
Work = K2 - K1,
where Work is the work done on the ball. Rearranging this equation, we have:
K2 = Work + K1.

6. Now we can substitute the values from the problem into the equation to find K2:
K2 = 86.6 J + K1.

7. Finally, use the equation for kinetic energy to convert K2 into velocity. Rearranging the equation, we get:
K2 = 0.5 * m * v2^2.
Solving for v2, we have:
v2 = sqrt((2 * K2) / m).

By following these steps and plugging in the given values, you can calculate the speed of the ball after it leaves the bat and is 22.5 m above the point of impact.