On Planet \( X \), a rock is thrown upward with an initial volocity of \( 16 \mathrm{~m} / \mathrm{s} \) from a plateau 26 moters high. Assume the acceleration due to gravity on Planet \( X \) is \( 6.6 \mathrm{~m} / \mathrm{s}^{2} \) (downward toward the planet.) Complete parts a. and b. (6) \( =|| | \mathrm{m} / \mathrm{s} \) ।

To analyze the motion of the rock thrown upward on Planet \( X \), we can use the kinematic equations of motion. The initial conditions are as follows: - Initial velocity (\( v_0 \)) = \( 16 \, \mathrm{m/s} \) (upward) - Initial height (\( h_0 \)) = \( 26 \, \mathrm{m} \) - Acceleration due to gravity (\( g \)) = \( -6.6 \, \mathrm{m/s}^2 \) (downward) ### Part a: Find the maximum height reached by the rock. To find the maximum height, we first need to determine the time it takes for the rock to reach its peak. At the maximum height, the final velocity (\( v \)) will be \( 0 \, \mathrm{m/s} \). Using the equation: \[ v = v_0 + at \] Setting \( v = 0 \): \[ 0 = 16 \, \mathrm{m/s} - 6.6 \, \mathrm{m/s}^2 \cdot t \] Solving for \( t \): \[ 6.6 \, t = 16 \] \[ t = \frac{16}{6.6} \approx 2.42 \, \mathrm{s} \] Now, we can find the maximum height using the equation: \[ h = h_0 + v_0 t + \frac{1}{2} a t^2 \] Substituting the known values: \[ h = 26 \, \mathrm{m} + 16 \, \mathrm{m/s} \cdot 2.42 \, \mathrm{s} + \frac{1}{2} (-6.6 \, \mathrm{m/s}^2) (2.42 \, \mathrm{s})^2 \] Calculating each term: 1. \( 16 \cdot 2.42 \approx 38.72 \, \mathrm{m} \) 2. \( \frac{1}{2} \cdot (-6.6) \cdot (2.42)^2 \approx -19.29 \, \mathrm{m} \) Now, substituting these values back into the height equation: \[ h \approx 26 + 38.72 - 19.29 \approx 45.43 \, \mathrm{m} \] Thus, the maximum height reached by the rock is approximately \( 45.43 \, \mathrm{m} \). ### Part b: Find the total time of flight until the rock hits the ground. To find the total time of flight, we need to determine when the rock returns to the height of \( 0 \, \mathrm{m} \). We can use the following kinematic equation: \[ h = h_0 + v_0 t + \frac{1}{2} a t^2 \] Setting \( h = 0 \): \[ 0 = 26 + 16t - 3.3t^2 \] Rearranging gives us: \[ 3.3t^2 - 16t - 26 = 0 \] Now we can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): - \( a = 3.3 \) - \( b = -16 \) - \( c = -26 \) Calculating the discriminant: \[ b^2 - 4ac = (-16)^2 - 4 \cdot 3.3 \cdot (-26) = 256 + 343.2 = 599.2 \] Now substituting into the quadratic formula: \[ t = \frac{16 \pm \sqrt{599.2}}{2 \cdot 3.3} \] Calculating \( \sqrt{599.2} \approx 24.48 \): \[ t = \frac{16 \pm 24.48}{6.6} \] Calculating the two possible values for \( t \): 1. \( t_1 = \frac{16 + 24.48}{6.6} \approx \frac{40.48}{6.6} \approx 6.13 \, \mathrm{s} \) 2. \( t_2 = \frac{16 - 24.48}{6.6} \) (this will yield a negative time, which we discard) Thus, the total time of flight until the rock hits the ground is approximately \( 6.13 \, \mathrm{s} \).