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

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