Question

A model rocket is launched with an initial upward velocity of 156 ft/s. The rocket's height h (In feet) after t seconds is given by the following.
h=156t-16t²
Find all values of t for which the rocket's height is 60 feet.
Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
Explanation
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ground
t = 0 seconds
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Answer 1

The quadratic equation that gives the height of the rocket, h = 156·t - 16·t² is evaluated at h = 60 feet to give the two times the rocket's height is 60 feet as 0.40 seconds and 9.35 seconds.

What is a quadratic equation?

A quadratic equation is an equation of the second degree that can be expressed in the form; a·x² + b·x + c = 0, where the letters, a, and b represents the coefficients of x and c is a constant.

The initial velocity of the rocket = 156 ft./s upwards

The given equation of the rocket is: h = 156·t - 16·t²

The times when the rocket height is 60 feet are found by plugging in the value h = 60, in the equation of the vertical height of the rocket as follows:  

h = 60 = 156·t - 16·t²

156·t - 16·t² - 60 = 0

4·(39·t - 4·t² - 15) = 0

Therefore: $39\cdot t - 4\cdot t^2 - 15 = \dfrac{0}{4} =0$

39·t - 4·t² - 15 = 0

-4·t² + 39·t - 15 = 0

From the quadratic formula which is used to solve the quadratic equation of the form; f(x) = a·x² + b·x + c, is presented as follows;

$x = \dfrac{-b\pm\sqrt{b^2-4\cdot a \cdot c} }{2\cdot a}$

The solution of the equation, -4·t² + 39·t - 15 = 0, is therefore:

$t = \dfrac{-39\pm\sqrt{(39)^2-4\times (-4) \times (-15)} }{2\times (-4)}= \dfrac{-39\pm\sqrt{1281} }{-8}$

Therefore, when the height of the rocket is 60 feet, the times are: $t = \dfrac{-39-\sqrt{1281} }{-8}\approx 9.35$ and $t = \dfrac{-39+\sqrt{1281} }{-8}\approx 0.40$

The times when the height of the rocket is 60 feet, the times are:

t ≈ 9.35 s, and t ≈ 0.40 s

Learn more about quadratic equations in algebra here:

brainly.com/question/472337

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