- Define the height function \(h(t) = -t^2 + \frac{4}{3}t + \frac{1}{4}\).
- Set \(h(t) = 0\) to find the landing time, resulting in the quadratic equation \(t^2 - \frac{4}{3}t - \frac{1}{4} = 0\).
- Apply the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to identify the roots.
- Select the positive root as the landing time, which is \(t = 1.5\).
- The final answer is \(\boxed{1.5}\).
Explanation
- Setting up the Equation
To determine when the grasshopper lands on the ground, we need to find the time \(t > 0\) such that the height \(h(t)\) is equal to zero. The height is given by the function:
\[h(t) = -t^2 + \frac{4}{3}t + \frac{1}{4}\]
Setting \(h(t) = 0\), we have the quadratic equation:
\[-t^2 + \frac{4}{3}t + \frac{1}{4} = 0\]
Multiplying the entire equation by \(-1\) gives:
\[t^2 - \frac{4}{3}t - \frac{1}{4} = 0\]
- Applying the Quadratic Formula
We use the quadratic formula to solve for \(t\). For a quadratic equation in the form \(at^2 + bt + c = 0\), the roots are given by:
\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In our equation \(t^2 - \frac{4}{3}t - \frac{1}{4} = 0\), the coefficients are:
\(a = 1\)
\(b = -\frac{4}{3}\)
\(c = -\frac{1}{4}\)
Substituting these values into the formula, we get:
\[t = \frac{-(-\frac{4}{3}) \pm \sqrt{(-\frac{4}{3})^2 - 4(1)(-\frac{1}{4})}}{2(1)}\]
- Solving for Time
First, we calculate the term under the square root (the discriminant):
\[(-\frac{4}{3})^2 - 4(1)(-\frac{1}{4}) = \frac{16}{9} + 1 = \frac{16}{9} + \frac{9}{9} = \frac{25}{9}\]
Now, we substitute this back into our expression for \(t\):
\[t = \frac{\frac{4}{3} \pm \sqrt{\frac{25}{9}}}{2} = \frac{\frac{4}{3} \pm \frac{5}{3}}{2}\]
This gives us two potential values for \(t\):
- \(t_1 = \frac{\frac{4}{3} + \frac{5}{3}}{2} = \frac{\frac{9}{3}}{2} = \frac{3}{2} = 1.5\)
- \(t_2 = \frac{\frac{4}{3} - \frac{5}{3}}{2} = \frac{-\frac{1}{3}}{2} = -\frac{1}{6}\)
Since time must be positive (\(t > 0\)), we discard \(t_2 = -\frac{1}{6}\) and keep \(t_1 = 1.5\). Thus, the grasshopper lands on the ground after 1.5 seconds.
Examples
This type of quadratic modeling is common in physics and engineering to predict the flight path of projectiles, such as a ball thrown in the air, an arrow fired from a bow, or even a water fountain's arc. By determining the time when the height reaches zero, we can calculate the distance traveled (range) or plan safety measures by knowing exactly when an object will return to the ground. Understanding these trajectories is essential for sports science, architecture, and logistics.