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.

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  1. v = Vi - 9.81 t

    20 = Vi - 9.81 (2.1)

    solve for Vi


    Use that Vi to get v at t = 4.7

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