To solve this problem we will apply the concepts of acceleration according to the laws of kinetics. Acceleration can be defined as the change in velocity per unit time, that is,
[tex]a = \frac{v-u}{t}[/tex]
Here,
v = Final velocity
u = Initial velocity
t = Time
Rearranging to find the initial velocity,
[tex]u = v-at[/tex]
Now the acceleration is equal to the gravity, then,
[tex]u = v+gt[/tex]
[tex]u = 17+(9.8)(1.8)[/tex]
[tex]u = 34.64m/s[/tex]
The velocity of the shell at 5.5s after the launch is,
[tex]v = u+at[/tex]
[tex]v = u-gt[/tex]
[tex]v = 34.64-(9.8)(5.5)[/tex]
[tex]v = -19.26m/s[/tex]
Therefore the magnitude of the velocity of the shell at 5.5s is -19.26m/s
