Asked by Mike
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.
Answers
Answered by
Damon
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
20 = Vi - 9.81 (2.1)
solve for Vi
then
Use that Vi to get v at t = 4.7
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