An object is released from rest. How far does it fall during the third second of its fall?

To determine how far an object falls during the third second of its fall, we can use the equation for the distance traveled by an object under constant acceleration:

d = (1/2) * g * t^2,

where:
- d is the distance traveled,
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time in seconds.

In this case, the object is falling from rest, so its initial velocity (u) is 0 m/s. Therefore, the equation can be simplified to:

d = (1/2) * g * t^2.

Let's calculate the distance the object falls during the third second:

t = 3 seconds
g = 9.8 m/s^2

Plugging these values into the equation, we have:

d = (1/2) * 9.8 * (3^2)
= (1/2) * 9.8 * 9
= 44.1 meters.

Therefore, the object falls approximately 44.1 meters during the third second of its fall.

find the d istance fell at end of third, and subtract the distance at the end of second.

d= d(3)-d(2)=1/2 g (3^2-2^2)
check that.