On Planet \( X \), a rock is thrown upward with an initial velocity of \( 16 \mathrm{~m} / \mathrm{s} \) from a plateau 26 meters 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 . a. Find the velocity of the rock 5 seconds into its flight. Use a positive sign for upward velocity and a negative sign for downward velocity. \( v(5)=\square \)

To find the velocity of the rock 5 seconds into its flight, we can use the following kinematic equation for velocity: \[ v(t) = v_0 - g t \] where: - \( v(t) \) is the velocity at time \( t \), - \( v_0 \) is the initial velocity, - \( g \) is the acceleration due to gravity (which is positive in this context but acts downward, so we will subtract it), - \( t \) is the time in seconds. Given: - \( v_0 = 16 \, \mathrm{m/s} \) (upward), - \( g = 6.6 \, \mathrm{m/s}^2 \) (downward), - \( t = 5 \, \mathrm{s} \). Now, substituting the values into the equation: \[ v(5) = 16 \, \mathrm{m/s} - (6.6 \, \mathrm{m/s}^2)(5 \, \mathrm{s}) \] Calculating the second term: \[ 6.6 \, \mathrm{m/s}^2 \times 5 \, \mathrm{s} = 33 \, \mathrm{m/s} \] Now substituting this back into the velocity equation: \[ v(5) = 16 \, \mathrm{m/s} - 33 \, \mathrm{m/s} = -17 \, \mathrm{m/s} \] Thus, the velocity of the rock 5 seconds into its flight is: \[ v(5) = -17 \, \mathrm{m/s} \] This indicates that the rock is moving downward at 17 m/s at that time.