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 = 29.37metres (to 2 d.p.)
Alternatively, find "time" needed to fall and use that.
Since all objects accelerate at 9.8ms-2 regardless of mass, we know that this drop took 24/9.807 = 2.4472seconds (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 = 29.37metres (to 2 d.p.)
To quickly explain:
On average, it's fallen at (24 + 0)/2 = 12ms-1
So the distance fell was 12 * 2.4472 = 29.37metres (to 2 d.p.)