A cannon fires a shell straight upward; 1.6 s after it is launched, the shell is moving upward with a speed of 19 m/s. Assuming air resistance is negligible, find the speed (magnitude of velocity) of the shell at launch and 5.5 s after the launch.

-19.259m/s

Explanation:

Given;

Final velocity = 19m/s

time t = 1.6s

u is the initial velocity

g is the acceleration due to gravity = 9.81m/s²

Using the equation of motion to first get the initial velocity of the shell:

v = u-gt

19 = u - (9.81)(1.6)

19 = u - 15.696

u = 19+15.696

u = 34.696m/s

The initial velocity of the shell is 34.696m/s

Next is to find the speed of the shell 5.5s after the launch

Using the equation of motion:

v = u-gt

v = 34.696-9.81(5.5)

v = 34.696 - 53.955

v = -19.259m/s

The negative value of the velocity shows that the velocity is travelling in the downward direction

their crystal patterns have been disrupted by stress

he made 11 calls in one day