8 A 1 kg ball is thrown into the air with an initial velocity of \( 35 \mathrm{~m} / \mathrm{s} \). A) How much kinetic energy does the ball have? B) How much potential energy does it have when it reaches the top of it's ascent upward and temporarily stops before coming back down? You do not have to work. C) Use your answer from part B, the mass of the ball, and the acceleration due to gravity to help you solve for the height of the ball. (PE=mgh) \( \square \)

To find the kinetic energy (KE) of the ball when thrown, we use the formula \( \text{KE} = \frac{1}{2} mv^2 \). Plugging in the values, \[ \text{KE} = \frac{1}{2} (1\, \text{kg}) (35\, \text{m/s})^2 = \frac{1}{2} (1)(1225) = 612.5\, \text{J}. \] So, the ball has \( 612.5 \, \text{J} \) of kinetic energy when it's thrown. At the top of its ascent, the ball's potential energy (PE) can be calculated using the formula \( \text{PE} = mgh \). However, as mentioned, you don't need to work through the calculations. It temporarily stops, indicating that all the initial kinetic energy has been converted to potential energy. Therefore, \( PE = 612.5 \, \text{J} \). Now, to find the height \( h \), rearranging the formula gives us \( h = \frac{PE}{mg} \). Using \( PE = 612.5 \, \text{J} \), mass \( m = 1 \, \text{kg} \) and acceleration due to gravity \( g \approx 9.8 \, \text{m/s}^2 \): \[ h = \frac{612.5}{1 \times 9.8} \approx 62.5\, \text{m}. \] So, the ball reaches a height of about \( 62.5 \, \text{m} \) before coming back down.