If an object is dropped from a height of meters, the velocity (in ) at impact is given by v= √2gh , where g=9.8m/sec^2 is the acceleration due to gravity. Required:
a. Determine the impact velocity for an object dropped from a height of 10 m.
b. Determine the height required for an object to have an impact velocity of 26.8 m/sec (~ 60 mph). Round to the nearest tenth of a meter.

A.) V = 14 m/s

B.) h = 36.6 m

Explanation:

Given the formula v = √2gh

where g = 9.8m/sec^2 is the acceleration due to gravity.

A.) Determine the impact velocity for an object dropped from a height of 10 m.

Substitute height h in the given formula

V = √2gh 

V = √2 × 9.8 × 10

V = √196

V = 14 m/s

b. Determine the height required for an object to have an impact velocity of 26.8 m/sec (~ 60 mph). Round to the nearest tenth of a meter.

Substitute the velocity in the given formula and make height h the subject of formula.

26.8 = √2 × 9.8 × h

Square both sides

718.24 = 19.6h

h = 718.24 / 19.6

h = 36.64 m

h = 36.6 m

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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.