SOLUTION: Please help me and explain how to solve this problem. Thank you. The path of an arrow shot into the air is h=144t-16t^2 where h is the height in feet above the ground t seconds

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Question 702202: Please help me and explain how to solve this problem. Thank you.
The path of an arrow shot into the air is h=144t-16t^2 where h is the height in feet above the ground t seconds after it is released. Graph the equation to find what period of time the arrow is above 224 feet.
Answer by neatmath(302) About Me  (Show Source):
You can put this solution on YOUR website!

This equation represents a parabola,

where the vertex is at a maximum,

and it opens down because of the negative coefficient of t.

h=144t-16t%5E2

h=-16t%5E2%2B144t where a=-16, b=144, and c=0

We can easily find the vertex (h,k) if we know the proper formulae:

h=-b%2F2a

h=-144%2F%282%2A-16%29

h=9%2F2

k=f%28h%29

k=f%289%2F2%29

k=-16%289%2F2%29%5E2%2B144%2A%289%2F2%29

k=324

So our vertex is the point (9/2,324)

We can also find the t-intercepts,

by setting h=0 and solving for t.

In this case, our t-intercepts are 0 and 9.

graph%28+600%2C+600%2C+-1%2C+10%2C+-10%2C+350%2C+-16x%5E2%2B144x+%29

By inspection of the graph,

it looks like the period of time that the arrow is over 224 feet

is between approximately 2 seconds and 7 seconds.

This is about 5 total seconds.

Of course this is only an approximation! Our graphs are never exact.

To find the exact answer,

we would need to find the points where h equals 224 in our equation,

and then we could find the exact times.

We would do this by setting the equation equal to 224, and solving for t.

This would give us the extremes of the range of t-values where h was greater than 224.

PS It does indeed look like 2 and 7 are the exact t-values we need.

I hope this helps! Keep practicing! :)

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