A cargo of mass 50kg fall freely from rest at a height of 100m and came to rest having penetrated 5.0cm of the ground calculate the time taken to fall
calculate the velocity of the body when it hits the ground

A.) 4.52s

B.) 44.3 m/s

Explanation:

Given that a cargo of mass 50kg fall freely from rest at a height of 100m and came to rest having penetrated 5.0cm of the ground

The total height = 100 + 5/100

The total height = 100 + 0.05

The total height = 100.05 m

Since the cargo fall from rest, the initial velocity U will be equal to zero. That is,

U = 0

Use second equation of motion

h = Ut + 1/2gt^2

Substitutes g = 9.8 and others parameters into the formula

100.05 = 0 + 1/2 × 9.8 × t^2

100.05 = 4.9t^2

t^2 = 100.05/4.9

t^2 = 20.42

t = sqrt( 20.42)

t = 4.52 s

B.) To calculate the velocity of the body when it hits the ground,

You can achieve it by first calculating its kinetic energy. Or by using the first equation of motion.

So, let use first equation of motion

V = U + at

But U = 0

Substitute the time and g into the formula

V = 9.8 × 4.52

V = 44.29 m/s

Therefore, the velocity of the body when it hits the ground is 44.3 m/s approximately.

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.

answer is slate

hope this