Answer:
Cannonball will be in flight before it hits the ground for 2.02 seconds
Explanation:
Initial height from ground = 20 meter.
We have equation of motion , [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 [tex]m/s^2[/tex], we need to calculate time when s = 20 meter.
Substituting
[tex]20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds[/tex]
So it will take 2.02 seconds to reach ground.