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2 years ago

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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.

1 answer:

ivolga24 [154]2 years ago

3 0

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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