Problem solving and reasoning

Question

a flare is launched from the side of a cliff 72 m high. The flare follows the path given by the formula h= -8t²+ 56t +72 where h is the height in metres and t is the time in seconds. Find the time at which the flare is 104 m or higher.

Answer 1

h= -8t²+ 56t +72

 

This is a parabola opening downward when graphed.  It starts at (0, 72) and

goes upwards to the vertex height and then downward.  Thus, since the vertex

is higher than 104m it will pass through a height of 104m twice, once on the

way up and once on the way down.

 

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NOTE:  The (t, h) coordinate of the vertex is at 3.5 seconds and 170m which can

be worked out by  v = (-b/2a, f(-b/2a)) with a=-8, b = 56, c = 72

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At what two times is the height equal to 104m is your basic question.  The difference

between the two times will give you your answer.

 

-8t²+ 56t +72 = 104

-8t² + 56t - 32 = 0

divide out the -8 from each term

t² - 7t + 4 = 0

 

Using the quadratic equation with a=1, b=-7, c=4:

t = [7 ±√((-7)²-4(1)(4))]/2(1) = (7 ±√33)/2

 

From this we get t = 0.6277 seconds or t = 6.3723 seconds

 

It takes 0.6277 seconds to reach a height of 104m initially, and

it takes 6.3723 seconds to again be at 104m on the way down.

Total time the flare is at 104m or higher is:

 

6.3723 - 0.6277 = 5.7446 seconds