Answer:
Explanation:
Given
Cannon is fired with a velocity of [tex]u=72.50\ m/s[/tex]
Using Equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
where
[tex]y=displacement[/tex]
[tex]u=initial\ velocity[/tex]
[tex]a=acceleration[/tex]
[tex]t=time[/tex]
after time [tex]t=3.3 s[/tex]
[tex]y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2[/tex]
[tex]y=239.25-53.36[/tex]
[tex]y=185.89\ m[/tex]
So after 3.3 s cannon ball is at a height of 185.89 m
