Question

A ball is thrown into the air from the ground. The ball’s height over time can be modeled with a quadratic function. The table shows the time, t, in seconds, and the height of the ball, h, in feet.
Using the intercepts from the table, the factored form of the quadratic function can be written as f(t) = at(t – 4).
The quadratic function that models the scenario is f(t) =
t2 +
t.
After
seconds, the ball attains its maximum height of
feet.

Answer 1

Answer:

Throwing a ball

Suppose you throw a ball straight up from the ground with a velocity of 80 ft. per second. As the ball moves upward, gravity slows it. Eventually the ball begins to fall back to the ground. The height h of the ball after t seconds in the air is given by the quadratic function h(t) = -16t2 +80t. How high does the ball go? How many seconds does it take for the ball to hit the ground?

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Answer 2

Answer:

After 2 seconds the ball attains its maximum height of 6 feet.

Step-by-step explanation:

• f(t) represents the height of the object. Now f(t) will become maximum when f''(t) < 0. [f''(t) is the 2nd derivative of f(t)]
• Here two quadratic equations are given. 1st take the equation,

         f(t) = t² + t = t(t + 1)  and f(t) = at(t – 4)

        Taking ratio of the two equations, $\frac{f(t)}{f(t)} = \frac{t(t+1)}{at(t-4)}$

                                                                  at(t - 4) = t(t + 1)

                                                                    a = $\frac{t + 1}{t - 4}$ ......(1)

• Another equation is, f(t) = at(t – 4) = at² - 4at.

        Taking the 1st derivative, f'(t) = $\frac{d(at^2-4at)}{dt}$ = 2at - 4a

        For finding critical value we have to make  f'(t) = 0

                                                                           or, 2at - 4a = 0   or, t = 2 s

        Taking the 1st derivative, f"(t) = $\frac{d(f'(t))}{dt}$ = $\frac{d(2at - 4a)}{dt}$ = 2a

        Putting the value of t in equation (1) we get, a = -3/2

        f"(t) = 2($\frac{-3}{2}$) = -3 < 0

        So, f(t) is maximum when t= 2s.

         Now f(t) = at(t – 4)

                  f(2) = 2($\frac{-3}{2}$)(2-4) = 6 feet

    ∴ After 2 seconds the ball attains its maximum height of 6 feet.

     

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