What is the instantaneous velocity of a freely falling object 28 s after it is released from a position of rest? The acceleration of gravity is 9.8 m/s2 .

2)What is its average velocity during this 28 s interval?

3) How far will it fall during this time?

This is not deep material.

vf=g*t solve

avg= vf/2

h= avgvelocity*time

To answer these questions, we'll need to use the equations of motion for a freely falling object.

1) The instantaneous velocity of a freely falling object 28 seconds after it is released can be calculated using the equation:
v = u + at,
where:
- v is the final velocity (which is what we need to find)
- u is the initial velocity (which is 0 since the object is released from rest)
- a is the acceleration due to gravity (which is -9.8 m/s^2 since it acts downward)
- t is the time (which is 28 seconds in this case)

Substituting the values into the equation, we can calculate the instantaneous velocity:
v = 0 + (-9.8 m/s^2) * 28 s
v = -274.4 m/s

Therefore, the instantaneous velocity of the object 28 seconds after it is released is -274.4 m/s.

2) The average velocity of the object during the 28-second interval can be calculated using the formula:
average velocity = (final velocity - initial velocity) / time

Since the object starts from rest, the initial velocity is 0. The final velocity is calculated in the previous step as -274.4 m/s. The time is 28 seconds. Substituting these values into the equation:
average velocity = (-274.4 m/s - 0) / 28 s
average velocity = -9.8 m/s

Therefore, the average velocity of the object during the 28-second interval is -9.8 m/s.

3) To calculate the distance the object will fall during this time, we'll use the equation for displacement:
s = ut + (1/2)at^2,
where:
- s is the displacement (which is what we need to find)
- u is the initial velocity (which is 0 since the object is released from rest)
- a is the acceleration due to gravity (which is -9.8 m/s^2 since it acts downward)
- t is the time (which is 28 seconds in this case)

Substituting the values into the equation, we can calculate the displacement:
s = 0 + (1/2) * (-9.8 m/s^2) * (28 s)^2
s = 1/2 * (-9.8 m/s^2) * 784 s^2
s = -3841.6 m

Therefore, the object will fall a distance of -3841.6 m during this time. Note that the negative sign indicates that the object is moving downwards.