In an archer shoots an arrow straight upward with an initial velocity of 192 ft/

Question

In an archer shoots an arrow straight upward with an initial velocity of 192 ft/sec form a height of 5 ft, them its height shown the ground in the feet at time 1 in seconds in given by the function h(t) = -10t^2 + 192t + 6. a. What is the maximum height reached by the arrow? b. How long does it take for the arrow to reach the ground? a. The maximum height it take for the arrow to reach the ground? (Simplify your answer) b. If tables seconds for the arrow to reach the ground. (Round to two decimal places as needed)

Explanation / Answer

h(t) = -16t^2 + 192 t + 5

maximum height occurs at the vertex

at t = -192 / 2(-16) = 6

maximum height is

-16(6)^2 +192(6) + 5 = 581

maximum height = 581 feet

time to reach the ground

at ground height = 0

plugging h(t) = 0 and solving for t

0 = -16t^2 + 192 t + 5

t = { -192 +- sqrt (192^2 - 4(-16)(5) } / -32

t = 12.026

it takes 12.03 seconds to reach the ground

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