Question

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
=
−
16
x
2
+
119
x
+
57
y=−16x
2
+119x+57

Answer 1

Explanation:

the target height (ground) is 0 ft, I assume.

the rocket starts from the height of the tower.

FYI - this we get for x = 0 seconds, which calculates the height of the rocket before it even starts.

y = -16x² + 119x + 57

for x = 0 we get y = 57 ft

anyway, we need to solve

0 = - 16x² + 119x + 57

the general solution for a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = -16

b = 119

c = 57

x = (-119 ± sqrt(119² - 4×-16×57))/(2×-16) =

= (-119 ± sqrt(14161 + 3648))/-32 =

= (-119 ± sqrt(17809))/-32

x1 = (-119 + 133.4503653...)/-32 = -0.451573916...

x2 = (-119 - 133.4503653...)/-32 = 7.889073916...

the negative solution as time does not make sense (this would be the time calculated back to the ground at the start), so x2 is our only solution :

the rocket will hit the ground after 7.89 seconds.