Final answer:
The cannon shell fired straight up with an initial velocity of 300 m/s lands back on the ground after approximately 122.4 seconds, following the rules of projectile motion and neglecting air resistance.
Step-by-step explanation:
The student is asking how long it will take for a cannon shell that is shot straight up with an initial upward velocity of 300 m/s to land back on the ground, assuming no air resistance. The problem is one of projectile motion and can be solved using basic kinematic equations. To find the time it will take for the shell to reach the ground again, you can use the following kinematic equation for motion under constant acceleration (which, in this case, is due to gravity):
final position = initial position + (initial velocity × time) + (½ × acceleration × time²)
Since the final position is the same as the initial position (ground level) and the acceleration due to gravity is -9.8 m/s² (negative because it's opposite the initial upward direction), the equation simplifies to:
0 = 0 + (300 m/s × time) - (½ × 9.8 m/s² × time²)
Solving this quadratic equation, we find that time is approximately 61.2 seconds for the shell to reach its peak and return to the ground, meaning the total time spent in the air is twice this: about 122.4 seconds.