You throw a 3.00-N rock vertically into the air from ground level. You observe that when it is 14.9 m above the ground, it is traveling at 24.9 m/s upward. Part A Use the work-energy theorem to find the rock's speed just as it left the ground.

rock's speed just as it left the ground = 30.2 m/s

Explanation:

The work energy theorem explains that the change in kinetic energy between two points = workdone between the two points.

Mass of the rock = 3/9.8 = 0.306 kg

Kinetic energy as it leaves the ground = mv₀²/2 = 0.153 v₀²

Kinetic energy at the height it reached = m(24.9²)/2 = 0.306 × 24.9²/2 = 94.862 J

Work done by the force of gravity in between those two points = (-mgh) = - 3 × 14.9 = - 44.7 J

94.862 - 0.153 v₀² = - 44.7

0.153 v₀² = 94.862 + 44.7 = 139.562

v₀ = 30.2 m/s

double the distance = divide strength by 4

your answer is, orbit around the nucleus