A diver running 1.8m/s dives out horizontally from the edge of a vertical cliff and 3.0s later reaches the water below. How high was the cliff, and how far from its base did the diver hit the water?

Answer 1

V initial for x axis: 1.8
V initial for y axis: 0.0
Acceleration on x axis: 0
Acceleration on y axis:-g=-9.81
Δ time: 3
Y final:0
X initial: 0
looking for: Y initial and X final

eqt 2-11b: Δx=Vi(Δtime)+(a*Δt^2)/2

for x axis:
Δx=Vi(Δt)+ zero (canceled out)
= 1.8(3)=5.4
5.4 m far from base

for y axis:
Δy= zero (canceled)+(a*Δt^2)/2
=(-9.81*3^2)/2
=-44.1
-44.1m is your height of cliff