Question

A model rocket is launched with an intitial upupward velocity of 202 ft/s. The rocket’s height h (in feet) after seconds is given by the following.
H=202t-16t^2

Answer 1

Given that the height of a object thrown within a gravitational field is given as a quadratic equation in time, t, the time at which the object is at a specified height can be found by the quadratic formula

The values of, t, for which the height of the rocket is 82 feet are;

t = 12.21 seconds or t = 0.42 seconds

Question: Parts of the question that appear missing but can be found online are (i) To find the values of the time, t, when the rocket's height is 82 feet and round answers to the nearest hundredth

The known parameters of the rocket are;

The initial upward velocity of the rocket, v = 202 ft./s

The given function representing the height, h, (in feet) after t seconds is presented as follows;

H = 202·t - 16·t²

The unknown;

The time for which the height of the rocket is 82 feet

Method;

Substitute H = 82 feet in the height function equation and solve for t as follows;

When H = 82, we get;

82 = 202·t - 16·t²

Therefore;

16·t² - 202·t + 82 = 0

Dividing the above equation by 2 gives;

(16·t² - 202·t + 82)/2 = 0/2

8·t² - 101·t + 41 = 0

By using the quadratic formula, we get;

$\mathbf{t = \dfrac{101 \pm \sqrt{(-101)^2 - 4 \times 8 \times 41} }{2 \times 8} = \dfrac{101 \pm \sqrt{8889} }{16}}$

Therefore, the values of t given by rounding off to the nearest hundredth are;

t = 12.21 or t = 0.42

The values of the time, t, at which the height of the rocket is 82 feet are t = 12.21 seconds or t = 0.42 seconds

Learn more about equation models of height as a function of time here;

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