John and Tracy look from their high-rise balcony, which is 80m above
the ground, to a swimming pool below- not exactly below, but rather
20m out from the bottom of their building.  they wonder how fast would
they have to jump horizontally to succeed in reaching the pool.  What
is the answer?

this is actually relatively simple to calculate using the equation
s=VoT+1/2aT^2
where s is the distance and Vo is intial velocity, T is time and a is
acelleration due to gravity
plug in 80m and 0 for intial velocity and -9.8m/s^2 for acceleration
and solve for time
T=4.04seconds
then use s=VT
to solve for velocity using 20m for distance and 4.04 for time and solve
V=4.95m/s

HW question, right?
Gregorious answer is basically correct except for 1 typo (-80 instead
of 80 below), but you may not be able to follow his solution fully.
Here's some clarifications to make Gregorious answer clearer. The
changes are in BOLD to help you understand.

this is actually relatively simple to calculate using the equation
s=VoT+1/2aT^2
where s is the VERTICAL distance and Vo is intial VERTICAL velocity, T
is time and a is acelleration due to gravity
plug in -80m and 0 for s AND intial VERTICAL velocity and -9.8m/s^2
for acceleration
and solve for time
T=4.04seconds
then use s=VT WHERE S IS THE HORIZONTAL DISTANCE AND V IS THE
HORIZONTAL VELOCITY to solve for THE HORIZONTAL velocity using 20m for
HORIZONTAL distance and 4.04 for time and solve
V=4.95m/s (HORIZONTAL VELOCITY)