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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 = -16x2 + 247x + 141

Respuesta :

Answer: 15,99s

when the rocket hit the ground => y = 0

=> -16x² + 247x + 141 = 0 (*)

a = - 16      b = 247     c = 141

Δ = b² - 4ac = 247² + 16.4.141 = 70033 > 0 => (*) has 2 solutions

=> x = [tex]\frac{247-\sqrt{70033} }{32}=-0,55(removed)[/tex]

or x = [tex]\frac{247+\sqrt{70033} }{32}=15,99(s)[/tex]

Step-by-step explanation:

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