You want to know how tall a cliff is, so you drop a rock off the edge. You estimate that the rock has a mass of about 15 kg. You hear a splash 1.00 seconds later. How tall is the
cliff?

Analysing the Question:

Neglecting air resistance

We are given:

mass (m) = 15 kg

time taken by the rock to reach the water (t) = 1 second

height of the cliff (h) = h m

initial velocity of the rock (u) = 0 m/s  [ the rock was 'dropped' and not 'thrown' downwards]

acceleration of the rock due to gravity (a) = 9.8 m/s²

Finding the height to the cliff:

from the seconds equation of motion

s = ut + 1/2*at²

replacing the variables

h = (0)(1) + 1/2 * (9.8) *(1)²

h = 4.9 m

Therefore, the cliff is 4.9 m high

your answer would be fusion

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer