1. ## [SOLVED] vertical motion

The height of an object moving vertically is given by:

s=-16t^2+96t+112

with s in ft and t in sec. Find a) the object's velocity when t=0 b) its maximum height and when it occurs, and c) its velocity when s=0

I don't even know where to start

2. Don't treat it as a physics question.

Find the vertex ( $\frac{-b}{2a}$)
Then plug in two random numbers (neg and pos) to find the other two points and graph.

(t is just x so look at the x axis for t)
t=0 is when the parabola hits the y axis. (I.e. Just set everything multiplied by t to 0)
max height is just the vertex.
(and for velocity at s=0 just set the equation equal to =0)sorry I thought v=s

3. well i have to figure it using derivatives

4. use the delta process and plug in for the resulting equation 0

velocity as t=0 is

$V=96x$
$v=96t+112$

5. so a. 96ft/s b. 256ft at t=3 and c. -128ft/sec?

6. so a. 96ft/s yes b. 256ft at t=3 yes and c. -128ft/sec? took me like 20 min to confirm but yes
.

7. Thank you very much for your time (twenty minutes and i might have quit if i were you) and effort. I finally get the problem... yea!!!