# polynomial Inequalities

• Oct 4th 2013, 07:30 AM
darkangel06
polynomial Inequalities
The function s(t)= -1/2gt2 +v0t +So can be used to calculate s, the height above a planet's surface in meters, where g is the acceleration due to gravity t is the time in seconds,Vo is the initial velocity in meters per second and so is the initial height in meters. The acceleration due to gravity on mars is g=-3.92 m/s2. Find to two decimal places how long it takes an object to hit the surface of mars if the object is dropped from 1000m above the surface.
• Oct 4th 2013, 07:42 AM
ebaines
Re: polynomial Inequalities
I don't understand why you titled this thread "polynomial inequalities." From the data you have been given you have $s=0m,\ v_o=0 m/s, \ s_0=2000m$. So the formula becomes:

$0m = -\frac 1 2 (3.92\frac m {s^2} ) t^2 + 0t +1000m$

Rearrange and solve for t.
• Oct 4th 2013, 07:55 AM
darkangel06
Re: polynomial Inequalities
because thats what it is :/
• Oct 4th 2013, 07:59 AM
darkangel06
Re: polynomial Inequalities
Ok if i solve this using the quadratic formula it is unsolvable
• Oct 4th 2013, 08:07 AM
ebaines
Re: polynomial Inequalities
Quote:

Originally Posted by darkangel06
Ok if i solve this using the quadratic formula it is unsolvable

This is simpler than you think, and it's not an inequality:

$0 = - \frac 1 2 (3.92 \frac m {s^2})t^2 +1000$

$\frac 1 2 (3.92 \frac m {s^2})t^2 = 1000$

$t^2 = \frac { 1000 m} {\frac 1 2 (3.92 \frac m {s^2})}$

$t = \sqrt {\frac {2 (1000m)}{3.92 \frac m {s^2}}}$

You can use the quadratic formula if you want to find the roots of $at^2 +bt+c = 0$, where $a= - \frac 1 2 (3.92), \ b=0,\ c = 1000$. Don't forget that minus sign in the 'a' coefficient!
• Oct 4th 2013, 08:42 AM
darkangel06
Re: polynomial Inequalities
well g= -3.92 not +3.92
• Oct 4th 2013, 10:26 AM
votan
Re: polynomial Inequalities
Quote:

Originally Posted by darkangel06
well g= -3.92 not +3.92

if a is acceleration, you would write s = at^2
if the acceleration is negatinve, as in the gravity, a = -3.92.