Answer:
The highest point of the cannonball displacement is h=37.06m, the time it uses to get there is t=2.75s, and its initial speed is [tex]v_i[/tex]=26.95m/s.
Step-by-step explanation:
The time the cannonball uses to return is the same it uses to get to the highest point of its displacement, so the time for each trip is 5.5s/2=2.75s
Initial speed [tex]v_i[/tex] can be determined by:
[tex]v_f -v_i =-g(t_f -t_i )[/tex]
[tex]0-v_i=-9.8(2.75-0)[/tex]
[tex]v_i=26.95m/s[/tex]
The height the cannonball reached is:
[tex]h=h_i+v_i t-\frac{1}{2}gt^2=0+26.95*2.75-\frac{1}{2}9.8(2.75)^2[/tex]
[tex]h=37.06m[/tex]
