math
posted by Natalie .
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.

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 = 3719.6
or 17.4 meters above ground
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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. 
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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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Thank you.