Final answer:
The student's question involves using the principles of projectile motion to determine the projectile's horizontal and vertical positions and its angle after 8 seconds of flight. The horizontal position is found using the horizontal component of initial velocity times time. The vertical position accounts for the initial vertical velocity and gravitational acceleration, and the angle is found using the arctangent of the velocity components.
Step-by-step explanation:
The student's question involves calculating the position and angle of a projectile 8 seconds after it has been fired from an artillery piece, which requires an application of the principles of projectile motion. Given the initial velocity of 200 m/s and launch angle of 15° with respect to the horizontal, we can use the equations of motion to find out the horizontal and vertical components of the projectile's position after 8 seconds.
To calculate the horizontal position (x), we use the equation x = Vx * t, where Vx is the horizontal component of the initial velocity and t is the time. Since Vx = V * cos(θ), where V is the initial velocity and θ is the launch angle, we find that Vx = 200 m/s * cos(15°). The horizontal position after 8 seconds is therefore this value multiplied by 8 seconds.
The vertical position (y) involves the initial vertical velocity component Vy and the effect of gravity. The vertical component Vy = V * sin(θ), and the vertical position y is given by y = Vy * t - 0.5 * g * t², where g is the acceleration due to gravity (9.81 m/s²). To find the angle at which the projectile will be traveling at t=8 seconds, we would need to calculate the new vertical velocity Vy at that time and use arctan(Vy/Vx) to find the angle.