Posted by **Natalie** on Saturday, June 14, 2008 at 3:24pm.

I am trying to solve a problem using Newton's equation, which is (-1/2)(9.8)t^2+ vot+ h. The problem says that a car rolls of a 120 foot high cliff. I know how to right the equation for this problem which would be (-1/2)(9.8 m/s) t^2 + vot + 37m. I am trying to find out how high the car will befter 2 seconds.

- math -
**Damon**, Saturday, June 14, 2008 at 5:37pm
well, when it starts off the cliff it may have horizontal velocity but has no vertical velocity so Vo = 0

(1/2) * 9.8 = 4.9 so we might write that as

z = -4.9 t^2 + 0 + 37

so after two seconds we have:

z = -4.9 * 4 + 37

z = 37-19.6

or 17.4 meters above ground

==================================

by the way, the vertical velocity would be:

V = Vo - 4.9 t

V = 0 - 9.8

or 9.8 meters per second down

The horizontal velocity would be whatever horizontal velocity it started with because there is no acceleration in the horizontal plane, only vertical.

- whoops, correction -
**Damon**, Saturday, June 14, 2008 at 5:39pm
*

by the way, the vertical velocity would be:

V = Vo - 9.8 t

V = 0 - 19.6

or 19.6 meters per second down

The horizontal velocity would be whatever horizontal velocity it started with because there is no acceleration in the horizontal plane, only vertical.

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- whoops, correction -
**Natalie**, Saturday, June 14, 2008 at 5:54pm
Thank you.

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