Question

An object is dropped off a cliff . Assume there is no air resistance, so it accelerates at -9.8m/sec/sec what is the velocity at the end of 2 seconds?

Answer 1

The velocity of an object dropped off a cliff can be calculated using the equation:

v = v0 + at

Given:

v0 = 0 m/s (initial velocity)

a = -9.8 m/s^2 (acceleration due to gravity)

At the end of 2 seconds:

t = 2 s

v = v0 + at = 0 m/s + (-9.8 m/s^2) * 2 s = -19.6 m/s

The velocity of the object at the end of 2 seconds is -19.6 m/s, downward.

At the end of 5 seconds:

t = 5 s

v = v0 + at = 0 m/s + (-9.8 m/s^2) * 5 s = -49 m/s

The velocity of the object at the end of 5 seconds is -49 m/s, downward.

At the end of 12 seconds:

t = 12 s

v = v0 + at = 0 m/s + (-9.8 m/s^2) * 12 s = -117.6 m/s

The velocity of the object at the end of 12 seconds is -117.6 m/s, downward.

It's worth mentioning that in the case of an object falling freely under the influence of gravity alone, the velocity will continue to increase