A boy on the edge of a vertical cliff 20.0 m high throws a stone horizontally outwards with a speed of 8.00 m/s.?

During the time that the stone is falling 20 meters, its vertical velocity increases at the rate of 9.8 m/s each second. To determine the time, use the following equation.

d = vi * t + ½ * a * t^2
vi is the initial vertical velocity. Since the stone is thrown horizontally, this is 0 m/s.

20 = ½ * 9.8 * t^2
t = √(20/4.9)
This is approximately 2.02 seconds.
Since the stone is thrown horizontally, use this time in following equation to determine the horizontal distance from the cliff does it strikes the ground.

d = v * t, v = 8 m/s
d = 8 * √(20/4.9)
This is approximately 16.2 meters.

For vertical motion,
s=20m
a=10m/s^2(g)
u=0 m/s
from eq.
s=ut+1/2*at^2
we get,
t=2 sec ---------------(1)

For horizontal motion
a=0
u=8m/s
t=2sec -------------(from(1))

from eq.
s=ut
we get
s=16m

so HORIZONTAL dispalcement is 16m

Take g = 10 m/s²

Consider the vertical motion :
s = vₒt + (1/2)at²
20.0 = 0 + (1/2)(10)t²
Time taken, t = 2 s

Consider the horizontal motion :
Horizontal distance = 8.00 × 2 = 16.0 m

20.0m