A cannon fires a shell straight upward; 1.7 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming air resistance is negligible, a. find the speed (magnitude of velocity) of the shell at launch and 5.5 s after the launch

(b) 5.5 s after the launch

a) 0.323m/s b) 54.278 m/s

Explanation:

The shell is moving upward with a speed of 17m/s

1.7s after it was launch

Using equation of motion on a straight line,

Vfinal = Vintial + acceleration due to gravity * time

Where V represent the speed (velocity)

Vinital = V (at 1.7s) - 9.81 * 1.7

Vinital = 17 - 16.677

a) Vinital = 0.323m/s

b) using the same equation,

V (after 5.5s) = Vinital + 9.81*5.5s

Vafter5.5s = 0.323 + 53.955

V after 5.5s = 54.278m/s