Final answer:
The final velocity of the projectile as it leaves the cannon is approximately 71.8 m/s.
Step-by-step explanation:
To find the final velocity of the projectile as it leaves the cannon, we can use the equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity (which is 0 m/s since the projectile starts from rest), a is the acceleration (which is 22.3 m/s^2), and s is the distance (which is 114 m).
First, we rearrange the equation to solve for v:
v^2 = 2as
Then, substitute the given values:
v^2 = 2(22.3 m/s^2)(114 m)
v^2 = 5146.8 m^2/s^2
Taking the square root of both sides, we find:
v = sqrt(5146.8 m^2/s^2) ≈ 71.8 m/s
So, the final velocity of the projectile as it leaves the cannon is approximately 71.8 m/s.