SOLUTION: A mode rocket is launched with an initial velocity of 200ft/s. The height (h), in feet, of the rocket (t) seconds after the launch is given by h = -16t^2 +200t. How many seconds

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A mode rocket is launched with an initial velocity of 200ft/s. The height (h), in feet, of the rocket (t) seconds after the launch is given by h = -16t^2 +200t. How many seconds       Log On


   

Question 162782This question is from textbook
: A mode rocket is launched with an initial velocity of 200ft/s. The height (h), in feet, of the rocket (t) seconds after the launch is given by
h = -16t^2 +200t. How many seconds after the launch will the rocket be 300 ft above the ground? Round to the nearest hundredth of a second.
So far I have it set up as such but no sure if its correct.
200t-300= -16t^2 + 200t - 200t
help appreciated.
 This question is from textbook

Found 2 solutions by Earlsdon, gonzo:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Starting with the given equation:
h%28t%29+=+-16t%5E2%2B200t You need to find the value of t (time in seconds) when h (height in feet) = 300 feet above the ground.
A little analysis before solving reveals that, assuming the rocket's maximum height is greater than 300 feet, there will be two such times when the rocket's height is 300 feet...the first on the way up and the second on the way down.
Presumably, you would like the first value of t (on the way up) when h = 300 feet, and you would expect this to be less than the second value of t (on the way down).
h%28t%29+=+-16t%5E2%2B200t Substitute h = 300
300+=+-16t%5E2%2B200t Rearrange the equation as:
16t%5E2-200t%2B300+=+0 Solve this quadratic equation using the quadratic formula:t+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a where a = 16, b = -200, and c = 300
t+=+%28-%28-200%29%2B-sqrt%28%28-200%29%5E2-4%2816%29%28300%29%29%29%2F2%2816%29 Simplify.
t+=+%28200%2B-sqrt%2840000-19200%29%29%2F32
t+=+%28200%2B-sqrt%2820800%29%29%2F32
t+=+6.25%2B4.51 or t+=+6.25-4.51
t+=+10.76 or t+=+1.74
The model rocket will reach a height of 300 feet in 1.74 seconds on its way up and will again be at a height of 300 feet in 10.76 seconds on its way down.

Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
initial velocity = 200 feet / second.
h = height of the rocket in feet.
t = number of seconds after launch.
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h = -16t^2 + 200t
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the question is: in how many seconds will the rocket be 300 feet above the ground.
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since h = -16t^2 + 200t, then if we want h to be 300 feet above the ground, the equation becomes
300 = -16t^2 + 200t
subtracting 300 from both sides of the equation and it becomes
-16t^2 + 200t - 300 = 0
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can't really see which factors would work by looking at the equation, so i will use the quadratic formula.
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for the quadratic formula
a = -16
b = 200
c = -300
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b = 200
b^2 = 40000
4ac = -16 * 4 * -300 = 19200
2a = -32
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%28-b+%2B+sqrt+%28b%5E2-4%2Aa%2Ac%29%29%2F%282a%29 becomes
%28-200+%2B+sqrt+%2840000-19200%29%29%2F%28-32%29
which becomes
t = 1.733060906 seconds.
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%28-b+-+sqrt+%28b%5E2-4%2Aa%2Ac%29%29%2F%282a%29 becomes
%28-200+-+sqrt+%2840000-19200%29%29%2F%28-32%29
which becomes
t = 10.75693909
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t can be either 1.73.... or 10.75693.....
substituting in the original equation shows the following
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using t = 1.733060906:
300 = -16 * t^2 + 200t becomes
300 = -16*(1.733060906)^2 + 200 * (1.733060906)
this becomes 300 = 300
looks like 17.3... is a good solution.
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using t = 10.75693909
300 = -16 * (10.75693909)^2 + 200 * (10.75693909)
this becomes 300 = 300
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it looks like t could be either
1.733060906
or
10.75693909
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since the initial velocity is 200 feet / second, we can calculate how far the rocket traveled in either of those seconds assuming that velocity is stable until the rocket has at least reached 300 feet.
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taking 1.733..., the rocket will have traveled 200 feet per second * 1.733... seconds = 346.6 feet
in order for it to reach 300 feet it would have to have been aimed almost straight up.
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taking 10.756..., the rocket will have traveled 200 feet per second * 10.756... seconds = 2151 feet.
this allows a much more shallow trajectory for it to reach 300 feet.
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both answers are plausible depending on the trajectory of the rocket and the velocity being at least as fast as when it was initially launched.
answers are:
t can be either...
1.733060906
or
10.75693909
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