a 6 kg rock falls from a cliff just before it hits the ground its speed is 24 m/s from what height did it fall from?

U = 0 The initial speed

V = 24 The final speed

a = 9.807 The acceleration

 

Using these we have 2 main methods of finding the distance it's fallen (and hence the height of the cliff).

First we can use a "Suvat" equation (Named such for using the 5 values: U, V, a, s and t). I won't get into quite why this equation works.

V² = U² + 2as

Re-arranging it gives:

s = (V² - U²) / 2a

Put in the numbers we got...

s = (24² - 0²) / (2 * 9.807)

s = 576 / 19.614

s = 29.3667788314469257

s = 29.37_metres (to 2 d.p.)

 

Alternatively, find "time" needed to fall and use that.

Since all objects accelerate at 9.8_ms^-2 regardless of mass, we know that this drop took 24/9.807 = 2.4472_seconds (to 4 d.p.).

We can then put it through this other "suvat" equation:

s = [(U + V) / 2] * t

s = [(0 + 24) / 2] * 2.4472

s = 29.37_metres (to 2 d.p.)

 

To quickly explain:

On average, it's fallen at (24 + 0)/2 = 12_ms^-1

So the distance fell was 12 * 2.4472 = 29.37_metres (to 2 d.p.)