A passenger on a stranded lifeboat shoots two distress flares into the air. The first flare travels a maximum height 196 feet and lands in the water after 7 seconds. The height (feet) of the second flare above the water is given by f(t) = -16t(t-8), where t is time (seconds) since the flare was shot. Which flare travels higher? Which remains in the air longer?

The second flare travels higher, 256 feet, and remains in the air longer, 8 seconds, than the first flare

Step-by-step explanation:

The given parameters are;

The height of the first flare = 196 feet

The duration the first flare spends in the air = 7 seconds

The function representing the height of the second flair is f(t) = -16t(t - 8)

Where;

t = The time since the passenger shot the second flare

The time the flair stays i the air is given by the time it takes for the height, f(t),  to be 0 as follows;

f(t) = 0 = -16t(t - 8)

-16t(t - 8) = 0

t = 0, or t = 8

The time the second flare stays in the air = 8 seconds

The maximum height reached by the second flair is given by the finding the maximum point of the curve of the function as follows;

d(f(t))/dt = 0 = d(-16t(t - 8))/dt = -32t + 128 at the maximum point (the coefficient of t² is negative in the quadratic equation)

-32t + 128 = 0

128 = 32t

32t = 128

t = 128/32 = 4

t = 4 at maximum height and f(4) = -16 × 4 × (4 - 8) = 256 feet

The maximum height reached by the second flare f(4) = -16 × 4 × (4 - 8) = 256 feet.

Therefore, the second flare travels higher, 256 feet, and remains in the air longer, 8 seconds, than the first flare, with maximum height of 196 feet, and time in the air of 7 seconds.

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step-by-step explanation: