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

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.