Question

A rock falls from rest off a high cliff. How far has the rock fallen when its speed is 39.2 meters per second?

( plss I need help lol)

Answer 1

I can help you. v^2=2as
39.2^2 = 2 x 9.8 x s
S= 78.4 m

Answer 2

The rock fell as far as 78.4 meters.

Given the following data:

• Initial speed = 0 m/s (since the rock fell from rest).

• Final speed = 39.2 m/s

We know that the acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find how far (distance) has the rock fallen, we would use the third equation of motion;

$V^2 = U^2 + 2aS$

Where:

• V is the final speed.

• U is the initial speed.

• a is the acceleration.

• S is the distance covered.

Substituting the given parameters into the formula, we have;

$39.2^2 = 0^2 + 2(9.8)S\\\\1536.64 = 0 + 19.6S\\\\19.6S = 1536.64\\\\S = \frac{1536.64}{19.6}$

Distance, S = 78.4 meters.

Therefore, the rock fell as far as 78.4 meters.

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