Answered

The height of an arrow shot upward can be given by the formula $s=v_0 t -16 t^2$, where $v_0$ is the initial velocity and $t$ is time. How long does it take for the arrow to reach a height of 48 ft if it has an initial velocity of $96 ft / s$? Round to the nearest hundredth.



The equation that represents the problem is $48=96 t-16 t^2$.

Solve $16 t^2-96 t+48=0$.



Complete the square to write $16 t^2-96 t+48=0$ as □.



Solve $(t-3)^2=6$. The arrow is at a height of 48 ft after approximately □ s and after □ s.

Verified Answer :

  • We begin with the height formula \(s = 96t - 16t^2\) and set \(s = 48\), resulting in the quadratic equation \(16t^2 - 96t + 48 = 0\).
  • Simplifying the equation and completing the square leads to the form \((t - 3)^2 = 6\).
  • Solving for \(t\) gives \(t = 3 \pm \sqrt{6}\).
  • The arrow reaches a height of 48 ft at approximately \(\boxed{0.55}\) s and \(\boxed{5.45}\) s.

Explanation

  1. Problem Analysis and Initial Equation

The height of the arrow is modeled by the formula \(s = v_0 t - 16t^2\). We are given that the initial velocity \(v_0 = 96\) ft/s and we want to find the time \(t\) when the height \(s = 48\) ft. Substituting these values gives the equation:
\[48 = 96t - 16t^2.\]
To solve for \(t\), we first rearrange the equation into standard quadratic form:
\[16t^2 - 96t + 48 = 0.\]

  1. Completing the Square

To simplify the equation \(16t^2 - 96t + 48 = 0\), we divide every term by 16:
\[t^2 - 6t + 3 = 0.\]
Next, we isolate the variable terms on one side:
\[t^2 - 6t = -3.\]
To complete the square on the left side, we take half of the coefficient of \(t\), which is \(-6\), and square it: \((\frac{-6}{2})^2 = (-3)^2 = 9\). Adding 9 to both sides of the equation gives:
\[t^2 - 6t + 9 = -3 + 9.\]
This simplifies to:
\[(t - 3)^2 = 6.\]
So the completed square form is \(\boxed{(t - 3)^2 = 6}\).

  1. Solving for Time and Conclusion

Now we solve \((t - 3)^2 = 6\) by taking the square root of both sides:
\[t - 3 = \pm \sqrt{6}.\]
This leads to two possible values for \(t\):
\[t = 3 \pm \sqrt{6}.\]
Using the value \(\sqrt{6} \approx 2.44949\):
\(t_1 = 3 - 2.44949 \approx 0.55051\) s
\(t_2 = 3 + 2.44949 \approx 5.44949\) s
Rounding to the nearest hundredth, the arrow is at a height of 48 ft at approximately \(0.55\) s and \(5.45\) s. Thus, the arrow is at a height of 48 ft after \(\boxed{0.55}\) s and after \(\boxed{5.45}\) s.

Examples

This problem represents a classic application of quadratic equations in physics. When an object is projected into the air, its height as a function of time follows a parabolic path. By solving for specific heights, engineers and scientists can predict exactly when an object, like an arrow or a launched probe, will pass specific markers. Such calculations are fundamental in ballistics, sports trajectory analysis, and even in designing fair games where predicting the timing of moving objects is key.