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12 August, 13:25

A cannon fires a shell straight upward; 2.3 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) in meters per second of the shell at launch and 4.7 s after the launch.

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  1. 12 August, 13:30
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    Answer: The launch speed is 43m/s

    4.7secs after launch speed is 4m/s

    Explanation: To solve this we use the first equation of motion but in this case our acceleration would be - 10m/s² since we are going upwards against gravity (launch).

    Vf = Vi - a*t

    Where Vf is the final velocity after launch, Vi is the initial velocity at launch, t is time in secs then a is acceleration

    a. From the question

    t = 2.3secs

    Vf = 20m/s

    a = - 10m/s²

    Substituting into the above equation we have that,

    20 = Vi - 10 * 2.3

    20 = Vi - 23

    Vi = 20+23

    Vi = 43 m/s

    Which is the speed at launch.

    b. The magnitude of speed (Vf) 4.7 sec after launch is calculated as follows using same procedure but here Vi is 43m/s as calculated

    Vf = 43 - 10*4.7

    Vf = 43 - 47

    Vf = - 4m/s

    But since we are asked to find the magnitude we neglect the negative sign.
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