A ball-launching machine can deliver $65.0 \mathrm{~W}$ of power to one $184 \mathrm{~g}$ ball launched straight up during $0.500 \mathrm{~s}$ that the ball is in contact with the machine. If the ball starts from rest, what speed will it have $4.82 \mathrm{~m}$ above its initial position, where the ball and machine are no longer in contact? $24.7 \mathrm{~m} / \mathrm{s}$ $16.1 \mathrm{~m} / \mathrm{s}$ $5.96 \mathrm{~m} / \mathrm{s}$ $19.5 \mathrm{~m} / \mathrm{s}$

Step 1 ：The power delivered by the machine is given as \(65.0 \mathrm{~W}\).

Step 2 ：The mass of the ball is given as \(184 \mathrm{~g}\) or \(0.184 \mathrm{~kg}\).

Step 3 ：The time the ball is in contact with the machine is given as \(0.500 \mathrm{~s}\).

Step 4 ：The height above the initial position where the ball and machine are no longer in contact is given as \(4.82 \mathrm{~m}\).

Step 5 ：The acceleration due to gravity is \(9.81 \mathrm{~m/s^2}\).

Step 6 ：The work done by the machine is the power multiplied by the time, which is \(65.0 \mathrm{~W} \times 0.500 \mathrm{~s} = 32.5 \mathrm{~J}\).

Step 7 ：The potential energy of the ball at the height of \(4.82 \mathrm{~m}\) is the mass of the ball multiplied by the acceleration due to gravity and the height, which is \(0.184 \mathrm{~kg} \times 9.81 \mathrm{~m/s^2} \times 4.82 \mathrm{~m} = 8.7002928 \mathrm{~J}\).

Step 8 ：The kinetic energy of the ball is the work done by the machine minus the potential energy, which is \(32.5 \mathrm{~J} - 8.7002928 \mathrm{~J} = 23.7997072 \mathrm{~J}\).

Step 9 ：The speed of the ball is the square root of twice the kinetic energy divided by the mass of the ball, which is \(\sqrt{2 \times 23.7997072 \mathrm{~J} / 0.184 \mathrm{~kg}} = 16.083919595832896 \mathrm{~m/s}\).

Step 10 ：\(\boxed{16.1 \mathrm{~m/s}}\) is the speed of the ball \(4.82 \mathrm{~m}\) above its initial position, where the ball and machine are no longer in contact.