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, equals, minus, 16, x, squared, plus, 261, x, plus, 146
y=−16x
2
+261x+146
To find the time when the rocket hits the ground, we need to find when the height of the rocket is equal to 0 (since hitting the ground means the height is 0).
So we set y = 0 in the equation:
0 = -16x^2 + 261x + 146
Now we need to solve for x. This is a quadratic equation, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In our equation, a = -16, b = 261, and c = 146. Plugging in these values, we get:
x = (-261 ± √(261^2 - 4*(-16)*146)) / (2*(-16))
x = (-261 ± √(67825 + 11776)) / -32
x = (-261 ± √79501) / -32
Now we calculate the square root:
√79501 ≈ 281.71
So we get two possible values for x:
x = (-261 + 281.71) / -32 ≈ 0.62 seconds
x = (-261 - 281.71) / -32 ≈ 24.88 seconds
Since the rocket is launched from a tower, we ignore the negative value. Therefore, the rocket will hit the ground in approximately 0.62 seconds.