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
=26.95m/s.
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
can be determined by:
![v_f -v_i =-g(t_f -t_i )](https://img.qammunity.org/2020/formulas/mathematics/high-school/3ibfn0i2cz8524pywhohth5qivo8bme9yd.png)
![0-v_i=-9.8(2.75-0)](https://img.qammunity.org/2020/formulas/mathematics/high-school/5xc7wtnk7pnasbga66yzny26by4dpdcg5k.png)
![v_i=26.95m/s](https://img.qammunity.org/2020/formulas/mathematics/high-school/9avzdbbv90neux2afi7x5xhol5r1iopjk7.png)
The height the cannonball reached is:
![h=h_i+v_i t-(1)/(2)gt^2=0+26.95*2.75-(1)/(2)9.8(2.75)^2](https://img.qammunity.org/2020/formulas/mathematics/high-school/n1kpszlu5n9af6ovt2sxl9ori5h1r1386v.png)
![h=37.06m](https://img.qammunity.org/2020/formulas/mathematics/high-school/4ia4po3xep7hdrts5z2yyuxq3qnusvdzva.png)