SOLUTION: If an object is projected vertically upward from ground level with an initial velocity of 240 ft/sec, then its distance s above the ground after t seconds is given by s = &#8722;

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Question 993688: If an object is projected vertically upward from ground level with an initial velocity of 240 ft/sec, then its distance s above the ground after t seconds is given by s = −12t^2 + 240t.For what values of t will the object be more than 1188 feet above the ground?
Found 2 solutions by Boreal, josmiceli:
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website!
1188>-12t^2+240t
-12t^2+240t-1188<0
-t^2+20t-99<0
t^2-20t+99>0
(t-11)(t-9).0
at t>9 and <11 seconds.
at 9 seconds, -972+2160=1188
at 11 seconds, -1452+2640=1188

Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
You need to find:

First find:

Complete the square:

and also:

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The values of for which are:

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Check:
Let

This is not true, as it must be
Now check for:
, and the inequality
must also be NOT true

SOLUTION: If an object is projected vertically upward from ground level with an initial velocity of 240 ft/sec, then its distance s above the ground after t seconds is given by s = &#8722;

SOLUTION: If an object is projected vertically upward from ground level with an initial velocity of 240 ft/sec, then its distance s above the ground after t seconds is given by s = −