SOLUTION: THE HEIGHT OF THE ROCKET AFTER t SECONDS WHEN FIRED STRAIGHT UP WITH AN INITIAL SPEED OF 150FT PER SECOND FROM AN INITIAL HEIGHT OF 2 CAN BE MODELED BY THE FUNCTION: s(t)=-16t^2+1

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Question 745513: THE HEIGHT OF THE ROCKET AFTER t SECONDS WHEN FIRED STRAIGHT UP WITH AN INITIAL SPEED OF 150FT PER SECOND FROM AN INITIAL HEIGHT OF 2 CAN BE MODELED BY THE FUNCTION: s(t)=-16t^2+150t+2
WHEN WILL THE ROCKET BE 300ft?
WHAT IS THE MEANING OF THE Y INTERCEPT
WHEN WILL IT REACH THE GROUND?
I AM UNSURE OF HOW TO START THE PROBLEM. DO I USE THE QUADRATIC FORMULA?

Found 3 solutions by josmiceli, rothauserc, Alan3354:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+s%28t%29++=+-16t%5E2+%2B+150t+%2B+2+
+s%28t%29+ is the height of the rocket from the ground
If you want the height to be +300+ ft, then
+s%28t%29+=+300+
+300++=+-16t%5E2+%2B+150t+%2B+2+
+-16t%5E2+%2B+150t+-+298+=+0+
Divide through by +2+
+-8t%5E2+%2B+75t+-+149+=+0+
Use quadratic wquation
+t+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29+
+a+=+-8+
+b+=+75+
+c+=+-149+
+t+=+%28-75+%2B-+sqrt%28+75%5E2+-+4%2A%28-8%29%2A%28-149%29+%29%29+%2F+%282%2A%28-8%29%29+
+t+=+%28-75+%2B-+sqrt%28+5625+-+4768+%29%29+%2F+%28-16%29+
+t+=+%28-75+%2B-+sqrt%28+857+%29%29+%2F+%28-16%29+
+t+=+%28-75+%2B+29.275%29+%2F+%28-16%29+
+t+=+-45.725%2F%28-16%29+
+t+=+2.858+ sec
In 2.858 sec the height of the rocket will be 300 ft
------------------
I have to run now- check my math. The method
should be OK

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
Indeed we use the given equation
s(t)=-16t^2+150t+2
a) when will rocket reach 300 ft
300 = -16t^2 + 150t + 2
-16t^2 + 150t - 298 = 0
use quadratic formula to solve
t = -150 plus or minus sqrt(150^2 - 4(-16)-298) / (-16*2)
b) note if t = 0 then s = 2 which is the initial distance
c) when will it reach the ground
to answer this we use the first derivative which is
-32t+150 and
t <75/16
so the rocket has a maximum height at (75/16, s(75/16))
now we want to know when it will reach the ground, this is just
2 * 75/16 or 75/8 = 9.38 seconds
a nifty free math calculator program for windows is "speedcrunch" :-)


Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
THE HEIGHT OF THE ROCKET AFTER t SECONDS WHEN FIRED STRAIGHT UP WITH AN INITIAL SPEED OF 150FT PER SECOND FROM AN INITIAL HEIGHT OF 2 CAN BE MODELED BY THE FUNCTION: s(t)=-16t^2+150t+2
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This is not a rocket, it's a projectile.
Rockets have engines and thrust.