A pitcher throws a 0.107-kg baseball, and it approaches the bat at a speed of 31.7 m/s. The bat does 77.9 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 22.7 m above the point of impact.
To determine the speed of the ball after it leaves the bat and is 22.7 m above the point of impact, we can use the principle of conservation of energy.
1. First, let's calculate the potential energy of the ball when it is 22.7 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 (0.107 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (22.7 m).
PE = 0.107 kg * 9.8 m/s^2 * 22.7 m
PE = 24.392 J
2. The total mechanical energy of the ball before it is hit by the bat is the sum of its initial kinetic energy and the potential energy. Since no other external work is done on the ball, the total mechanical energy is conserved. Thus, the total mechanical energy after the ball is hit is equal to the work done by the bat (77.9 J) plus the initial kinetic energy of the ball.
Total mechanical energy = Work done by the bat + Initial kinetic energy
3. The initial kinetic energy of the ball (KE) can be calculated using the formula:
KE = 0.5 * m * v^2
where m is the mass of the ball (0.107 kg) and v is the initial velocity (31.7 m/s).
KE = 0.5 * 0.107 kg * (31.7 m/s)^2
KE = 55.843 J
4. Adding the initial kinetic energy to the work done by the bat:
Total mechanical energy = 77.9 J + 55.843 J
Total mechanical energy = 133.743 J
5. The total mechanical energy after the ball is hit is equal to the sum of the final kinetic energy (KE) and the potential energy (PE) when the ball is 22.7 m above the point of impact. Let's assume the final kinetic energy is KF.
Total mechanical energy = KF + PE
133.743 J = KF + 24.392 J
6. Now, we can solve for the final kinetic energy:
KF = 133.743 J - 24.392 J
KF = 109.351 J
7. Finally, we can determine the velocity (vF) at this height using the final kinetic energy (KF) equation:
KF = 0.5 * m * vF^2
Rearranging the equation:
vF^2 = (2 * KF) / m
vF = sqrt((2 * 109.351 J) / 0.107 kg)
vF = 44.552 m/s
Therefore, the speed of the ball after it leaves the bat and is 22.7 m above the point of impact is approximately 44.552 m/s.
To determine the speed of the ball after it leaves the bat, we can use the principle of conservation of energy. The work done on the ball by the bat will be converted into kinetic energy after the ball leaves the bat.
The initial kinetic energy (KE1) of the ball can be calculated using the equation:
KE1 = (1/2) * m * v1^2
where m is the mass of the ball and v1 is the initial speed of the ball before it is hit by the bat.
Given:
m = 0.107 kg
v1 = 31.7 m/s
Plugging in the values, we can calculate KE1:
KE1 = (1/2) * 0.107 kg * (31.7 m/s)^2 = 54.365 J
Now, as the ball reaches a height of 22.7 m above the point of impact, the potential energy (PE2) of the ball can be calculated using the equation:
PE2 = 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.
Given:
m = 0.107 kg
g = 9.8 m/s^2
h = 22.7 m
Plugging in the values, we can calculate PE2:
PE2 = 0.107 kg * 9.8 m/s^2 * 22.7 m = 25.2424 J
Now, according to the principle of conservation of energy, the total mechanical energy (E1) of the ball before it is hit should be equal to the total mechanical energy (E2) of the ball after it leaves the bat:
E1 = KE1
E2 = KE2 + PE2
Since we want to find the speed of the ball after it leaves the bat, we can rearrange the equations:
KE2 = E2 - PE2
Given:
E2 = 77.9 J
Plugging in the values, we can calculate KE2:
KE2 = 77.9 J - 25.2424 J = 52.6576 J
Now, we can use the equation for kinetic energy (KE2) to find the final speed (v2) of the ball:
KE2 = (1/2) * m * v2^2
Rearranging the equation:
v2 = √(2 * KE2 / m)
Plugging in the values, we can calculate v2:
v2 = √(2 * 52.6576 J / 0.107 kg) = 23.2 m/s
Therefore, the speed of the ball after it leaves the bat and is 22.7 m above the point of impact is approximately 23.2 m/s.