Question

A man stands on the edge of a cliff with hands outstretched horizontally
carrying identical pebbles, one in each hand. At time t= 0, he drops the first pebble from rest.When this pebble has fallen by 2.0 m, he releases the second pebble. What is the vertical distance between the pebbles at t= 5s?

(You may neglect air resistance and assume that neither pebble
has hit the ground at 5.0 s.)

Answer 1

Particle 1
v = g t
d = (1/2) g t^2
how long to fall 2 meters?
4 = 9.81 t^2
t = .408 seconds

Particle 2
v' = g (t-.408)
d' = (1/2) g t(t -.408)^2

at t = 5
d = (1/2)(9.81)(25)
d' = (1/2)(9.81)(4.59)^2
so calculate d'-d

Answer 2

t is not 0.408. t^2 is 0.408. and t=0.63 s for 2nd h=5s-0.63s=4.37s
h=1/2gt^2
h1=1/2x9.81x(5)^2=123 m
h2=1/2x9.81x(4.37)^2=95.48 m

h1-h2=27.145 m