Thursday

April 17, 2014

April 17, 2014

Posted by **john** on Thursday, September 9, 2010 at 10:02pm.

- physic -
**osvaldo**, Saturday, June 25, 2011 at 5:55pmV 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

**Related Questions**

physics - A diver running 1.8m/s dives out horizontally from the edge of a ...

Physics - a diver running 1.6m/s dives out horizontally from the edge of a ...

Physics - A diver running 1.1 m/s dives out horizontally from the edge of a ...

Physics - A diver running 1.2 m/s dives out horizontally from the edge of a ...

physics - A diver running 1.3 m/s dives out horizontally from the edge of a ...

physics - A diver running 1.6 m/s dives out horizontally from the edge of a ...

physics - A diver running 1.0 m/s dives out horizontally from the edge of a ...

Physics - A diver running 1.69 m/s dives out horizontally from the edge of a ...

Physics - A diver running 1.69 m/s dives out horizontally from the edge of a ...

physics - A diver running 2.4 m/s dives out horizontally from the edge of a ...