\begin{tabular}{l} L Save \& Exit Practice Lesson: 1.8 Quadratic Equations \\ \hline Question 6 of 15, Step 1 of 1 \\ A rock is thrown upward with a velocity of 16 meters per second from the top of a 49 meter high cliff, and it misses the cliff on the way back down. When will the rock \\ be 6 meters from ground level? Round your answer to two decimal places. \\ \hline\end{tabular}

To determine when the rock will be 6 meters from ground level, we can use the kinematic equation for the height of the rock as a function of time. The height \( h(t) \) of the rock at time \( t \) can be modeled by the equation: \[ h(t) = -4.9t^2 + v_0t + h_0 \] where: - \( v_0 \) is the initial velocity (16 m/s), - \( h_0 \) is the initial height (49 m), - \( -4.9 \) is the acceleration due to gravity (in meters per second squared, divided by 2). Substituting the values into the equation, we have: \[ h(t) = -4.9t^2 + 16t + 49 \] We want to find when the height \( h(t) \) is 6 meters from the ground level. Therefore, we set up the equation: \[ -4.9t^2 + 16t + 49 = 6 \] Subtracting 6 from both sides gives: \[ -4.9t^2 + 16t + 43 = 0 \] To make it easier to solve, we can multiply the entire equation by -1: \[ 4.9t^2 - 16t - 43 = 0 \] Now we can use the quadratic formula to solve for \( t \): \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 4.9 \), \( b = -16 \), and \( c = -43 \). Calculating the discriminant: \[ b^2 - 4ac = (-16)^2 - 4(4.9)(-43) = 256 + 841.6 = 1097.6 \] Now substituting into the quadratic formula: \[ t = \frac{16 \pm \sqrt{1097.6}}{2 \cdot 4.9} \] Calculating \( \sqrt{1097.6} \): \[ \sqrt{1097.6} \approx 33.12 \] Now substituting back into the formula: \[ t = \frac{16 \pm 33.12}{9.8} \] Calculating the two possible values for \( t \): 1. \( t = \frac{16 + 33.12}{9.8} \approx \frac{49.12}{9.8} \approx 5.01 \) 2. \( t = \frac{16 - 33.12}{9.8} \approx \frac{-17.12}{9.8} \approx -1.74 \) (not a valid solution since time cannot be negative) Thus, the only valid solution is: \[ t \approx 5.01 \text{ seconds} \] Rounding to two decimal places, the rock will be 6 meters from ground level at approximately: \[ \boxed{5.01} \text{ seconds} \]