A cannon fires a shell straight upward; 1.7 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 5.4 s after the launch.

Answer 1

v = Vi - 9.81 t

at 1.7 s
19 = Vi - 9.81 (1.7)
solve for Vi, launch speed

at 5.4 s
v = Vi - 9.81(5.4)