SOLUTION: The word problem I am having troubles with is as follows..... The height (h) (in feet) of a rocket at time (t) seconds after liftoffi sgiven by h=-16t^2+800t. (a). how long d

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Question 112413: The word problem I am having troubles with is as follows.....
The height (h) (in feet) of a rocket at time (t) seconds after liftoffi sgiven by h=-16t^2+800t.
(a). how long dos it take the rocket to reach 3200 feet?
(b). What is the masimum height of the rocket.
THe following is what I have tried to complete but am doing something totally wrong. Can you please help.
-16t^2+800t=3200
(-16t)^2+800t=3200
(-16t)(-16t)+800t=3200
256-16t-16t+2t+800t=3200
256+2t+800t=3200
256-256+2t=800t=3200+256
2t+800t=3456
Can you please walk me through this and tell me where I went totally wrong.
Thanks so much!!!!
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Your height function is a quadratic equation. You have to use the quadratic formula or factoring to solve it.
a) Find t when h=3200
h=-16t%5E2%2B800t=3200
-16t%5E2%2B800t=3200
-16t%5E2%2B800t-3200=0 Subtract 3200 from both sides.
t%5E2-50t%2B200=0 Divide both sides by (-16).
t=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ Quadratic formula.
where
a=1
b=-50
c=200
t=+%2850+%2B-+sqrt%28+50%5E2-4%2A1%2A200+%29%29%2F%282%2A1%29+
t=+%2850+%2B-+sqrt%28+2500-800+%29%29%2F%282%29+
t=+%2850+%2B-+sqrt%28+1700+%29%29%2F%282%29+
t=+%2850+%2B-+41.23%29%2F%282%29+
There are two answers.
t=+%2850+%2B+41.23%29%2F%282%29=+45.62
t=+%2850+-+41.23%29%2F%282%29=+4.39
The rocket starts at zero feet climbs to 3200 and beyond.
Then it reaches a maximum height and starts to descend.
The first time it is at 3200 feet, 4.39 seconds, is before the rocket has reached its maximum height.
The second time it is at 3200 feet, 45.62 seconds, is after it reached the maximum.
a.) t=4.39 seconds
Check the answer.
-16t%5E2%2B800t=3200
-16%284.39%29%5E2%2B800%284.39%29=3200
-16%2819.27%29%2B3512=3200
-308.3%2B3512=3200
-308.3%2B3512=3200
3203.7=3200
Close enough.
+graph%28+300%2C+300%2C+-10%2C+50%2C+-1000%2C+11000%2C+-16x%5E2%2B800x%2C3200%29+
b.) We can find the maximum two ways.
The equation is a parabola and we have two points at the same height.
The average of the times (4.39 and 45.62) will give us the time when the parabola has its maximum.
t%5Ba%5D=%284.39%2B45.62%29%2F2
t%5Ba%5D=25.0
h=-16t%5E2%2B800t=3200
h=-16%2825%29%5E2%2B800%2825%29
h=-16%28625%29%2B800%2825%29
h=-10000%2B20000
h=10000
The other way is to take the derivative of the height function wrt time and set it to zero to find minima/maxima.
h=-16t%5E2%2B800t=3200
dh%2Fdt=-32t%2B800
-32t%2B800=0
-32t=-800
t=25
As before h(25)=10000 feet.