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A quarterback throws a football, and the model $y=-16 x^2+35 x+6.5$ describes the height ($y$), in feet, of the football $x$ seconds after it is thrown. If the ball is not caught, how long will it be in the air?



$\begin{array}{l}

0=-16 x^2+35 x+6.5 \\

x=\frac{-35 \pm \sqrt{35^2-4(-16)(6.5)}}{2(-16)}

\end{array}$



Rounded to the nearest tenth, the solutions of the equation are -0.2 and 2.4.



Why can you eliminate the solution of -0.2 in the context of this problem? Choose two correct answers.

  • It does not make sense for time to be negative.
  • The ball cannot hit the ground before it is thrown.
  • The height of the ball must be positive.



To the nearest tenth of a second, the ball is in the air for _____ s.

Verified Answer :

  • The projectile motion is modeled by the quadratic equation \(-16x^2 + 35x + 6.5 = 0\).
  • Using the quadratic formula, the solutions are approximately \(-0.2\) and \(2.4\).
  • The solution \(-0.2\) is discarded because time cannot be negative and the ball cannot hit the ground before it is thrown.
  • The ball remains in the air for \(\boxed{2.4}\) s.

Explanation

  1. Analyzing the Problem and Solving the Equation

The height of a football in feet, \(y\), at time \(x\) in seconds after it is thrown is given by the quadratic equation \(y = -16x^2 + 35x + 6.5\). To find the time the ball is in the air, we need to determine when it hits the ground, which occurs when the height \(y\) is equal to \(0\). Setting the height to zero gives us:
\[ -16x^2 + 35x + 6.5 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = -16\), \(b = 35\), and \(c = 6.5\), we can find the two potential times when the ball is at ground level. Substituting these values into the formula, we obtain:
\[ x = \frac{-35 \pm \sqrt{35^2 - 4(-16)(6.5)}}{2(-16)} \]
Calculating the discriminant, we get \(35^2 - 4(-16)(6.5) = 1225 + 416 = 1641\). The square root of \(1641\) is approximately \(40.5\). Thus, the two solutions are approximately:
\[ x_1 \approx \frac{-35 + 40.5}{-32} \approx -0.17 \]
\[ x_2 \approx \frac{-35 - 40.5}{-32} \approx 2.36 \]
Rounding to the nearest tenth, we have the solutions \(-0.2\) and \(2.4\).

  1. Interpreting the Results in Context

We have two mathematical solutions for the time the ball hits the ground: \(x = -0.2\) and \(x = 2.4\). In the context of this physics problem, we must interpret these values physically. The variable \(x\) represents the time elapsed since the ball was thrown, which starts at \(x = 0\). Therefore, negative time values do not describe the motion of the ball after it is thrown. Specifically, we can discard the solution \(x = -0.2\) for two main reasons:

  1. It does not make sense for time to be negative in this scenario.
  2. The ball cannot hit the ground before it is thrown (the event begins at \(x=0\)).

The third option, 'The height of the ball must be positive', is not a valid reason to discard the negative value because height can indeed be positive at negative time values (mathematically, the parabola exists for \(x < 0\)). Thus, the only physically valid time the ball hits the ground is \(x \approx 2.4\).

  1. Conclusion

Since the ball is thrown at \(x = 0\) and hits the ground at approximately \(x = 2.4\) seconds, the total duration the ball is in the air is \(2.4\) seconds. We have determined the valid solution and justified why the negative result is extraneous. Thus, the ball is in the air for \(2.4\) s.

Examples

This quadratic modeling technique is frequently used in engineering and sports analytics to predict projectile trajectories. By identifying the roots of a quadratic equation, analysts can determine the landing point of an object or the peak height, allowing them to optimize performance or ensure safety in design. For instance, in building construction or urban planning, similar models help predict the arc of debris or the range of fire suppression systems.

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