A 0.6 kg apple falls from a tree. At 0.6 s , the apple reaches the ground at 0 m . What is the potential energy of the apple? \[ \begin{array}{r}E_{p}=[?] \mathrm{J} \\ E_{p}=m g h\end{array} \] For gravitational acceleration, use \( \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2} \). 0 J 30 J 60 J

To find the potential energy (Eₚ) of the apple before it falls, we can use the formula Eₚ = mgh, where m is the mass in kilograms, g is the acceleration due to gravity, and h is the height in meters. If the apple takes 0.6 seconds to hit the ground and assuming it falls freely, we can determine the height using the formula h = 0.5gt². Plugging in g = 10 m/s² and t = 0.6 s, we find h ≈ 1.8 m. Then, Eₚ = 0.6 kg x 10 m/s² x 1.8 m = 10.8 J. Interestingly, the potential energy is converted to kinetic energy as it falls, so the initial energy in the apple before the fall is around 10.8 J, which would be in between the options given. Additionally, remember that potential energy is highest when an object is at its peak height, and as it begins to fall, that energy transforms into kinetic energy, which is at its maximum just before impact. So the 0 J at ground level indicates that all potential energy has been converted into kinetic energy!