A1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s.

(a) what impulse acts on the ball during the contact?

(b) if the ball is in contact with the floor for 0.020 s, what is the magnitude of the average force on the floor from by the ball? neglect air resistance

Explanation:

Given that,

Mass of the ball, m = 1.2 kg

Initial speed of the ball, u = 10 m/s

Height of the floor from ground, h = 32 m

(a) Let v is the final speed of the ball. It can be calculated using the conservation of energy as :

v = -25.04 m/s (negative as it rebounds)

The impulse acting on the ball is equal to the change in momentum. It can be calculated as :

J = -42.048 kg-m/s

(b) Time of contact, t = 0.02 s

Let F is the average force on the floor from by the ball. Impulse acting on an object is given by :

F = 0.8409 N

Hence, this is the required solution.

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.