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

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- Jan 1st 2010, 11:31 AMsmartartbug[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

- Jan 1st 2010, 11:35 AMintegral
Don't treat it as a physics question.

Find the vertex ($\displaystyle \frac{-b}{2a}$)

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 (Giggle) - Jan 1st 2010, 11:37 AMsmartartbug
well i have to figure it using derivatives

- Jan 1st 2010, 11:41 AMintegral
use the delta process and plug in for the resulting equation 0 (Thinking)

velocity as t=0 is

$\displaystyle V=96x$

$\displaystyle v=96t+112$ - Jan 1st 2010, 12:08 PMsmartartbug
so a. 96ft/s b. 256ft at t=3 and c. -128ft/sec?

- Jan 1st 2010, 05:58 PMintegralQuote:

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

- Jan 1st 2010, 06:07 PMsmartartbug
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!!!(Rofl)