A cannon fires a shell straight upward; 2.1 s after it is launched, the shell is moving upward with a speed of 20 m/s. Assuming air resistance is negligible, find the speed (magnitude of velocity) of the shell at launch and 4.7 s after the launch.

Answer 1

v = Vi - 9.81 t

20 = Vi - 9.81 (2.1)

solve for Vi

then

Use that Vi to get v at t = 4.7