Posted by Natalie on .
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,
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
==================================
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,
*
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.
Answer this Question 
whoops, correction 
Natalie,
Thank you.