Answer:
The minimum initial speed of the cannon ball is approximately 99.018 m/s
Explanation:
The parameters of the cannon are;
The maximum horizontal distance reached, R = 3,279 ft. ≈ 999.4392 m
The maximum horizontal range, [tex]R_{max}[/tex], is given by the following formula;
[tex]R_{max} = \dfrac{u^2}{g}[/tex]
Where;
u = The initial speed
g = The acceleration due to gravity ≈ 9.81 m/s²
[tex]R_{max}[/tex] = 999.4392 m
We get;
u = √(g × [tex]R_{max}[/tex])
∴ u = √(9.81 m/s² × 999.4392 m) ≈ 99.018 m/s
The minimum initial speed of the cannon ball (just as it leaves the cannon) that is needed for it to reach this distance, in m/s, u ≈ 99.018 m/s.
