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A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second.
y=-16x^2+187x+93
equation

Answer :

abidemiokin

Answer:

12.17secs

Step-by-step explanation:

Given the equation that models the height as y=-16x^2+187x+93

The rocket will hit the groung at y = 0

The equation becomes;

0 =-16x^2+187x+93

16x^2-187x-93 = 0

Factorize

x = 187±√187²-4(16)(-93)/2(16)

x = 187±√34,969+5952/32

x = 187±√40,921/32

x = 187±202.29/32

x = 187+202.29/32

x = 389.29/32

x = 12.17secs

Hence the required time that t will take is 12.17secs

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