Why does a squirrel's velocity not reach 31.3 m/s when it jumps off a 50 m high cliff?

Question:

Why isn't the velocity of the squirrel 31.3 m/s when it hits the ground after jumping from a 50 m high cliff?

Answer:

The velocity of the squirrel when it hits the ground is not 31.3 m/s because the final velocity of a falling object is determined by both the height it falls from and the acceleration due to gravity.

Explanation:

When a squirrel jumps from a 50 m high cliff, its final velocity when it hits the ground is not 31.3 m/s, contrary to what may be expected. This discrepancy in velocity can be explained by the effect of gravity on the falling object.

It's important to note that the final velocity of a falling object is not solely dependent on the height it falls from. The acceleration due to gravity, denoted by 'g', plays a significant role in determining the velocity of the object when it reaches the ground. On Earth, the acceleration due to gravity is approximately 9.8 m/s².

The formula to calculate the final velocity of a falling object is v = sqrt(2gh), where:
- v is the final velocity
- g is the acceleration due to gravity
- h is the height from which the object falls

For a squirrel jumping from a 50 m high cliff, the calculation would be:
v = sqrt(2 * 9.8 m/s² * 50 m)
v = sqrt(980 m²/s²)
v = 31.3 m/s

Therefore, the velocity of the squirrel when it hits the ground would actually be 31.3 m/s, in accordance with the formula for the final velocity of a falling object. The acceleration due to gravity and final velocity are key concepts in understanding this phenomenon.