Final answer:
The velocity of the projectile after 12 seconds is -24.72 meters per second, indicating that it is on its way down after reaching the peak of its trajectory.
Step-by-step explanation:
The question asks about the velocity of a projectile after 12 seconds when it was shot upward from the Earth's surface with an initial velocity of 93 meters per second. The velocity of the projectile after a given time can be calculated using the equation of motion for a freely falling object under gravity. The general equation for velocity in a uniformly accelerated motion is v(t) = v0 - gt, where v(t) is the final velocity at time t, v0 is the initial velocity, g is the acceleration due to gravity (approximately 9.81 m/s²), and t is the time elapsed. Assuming that up is positive and the acceleration due to gravity is upward, the velocity of the projectile after 12 seconds can be calculated as follows:
v(12s) = 93 m/s - (9.81 m/s² × 12 s) = 93 m/s - 117.72 m/s = -24.72 m/s.
Therefore, the velocity of the projectile after 12 seconds is -24.72 meters per second, indicating that it is moving downward.