Answer:
Explanation:
Given
Cannon is fired with a velocity of 
Using Equation of motion

where




after time 



So after 3.3 s cannon ball is at a height of 185.89 m
Answer:
Common Examples of Imagery
Taste: The familiar tang of his grandmother's cranberry sauce reminded him of his youth. Sound: The concert was so loud that her ears rang for days afterward. Sight: The sunset was the most gorgeous they'd ever seen; the clouds were edged with pink and gold.
I hope it's helpful!
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According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.
To solve this problem we must apply the concept related to the conservation of energy theorem.
PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so



PART B) Replacing the values given as,




Therefore the speed of the masses would be 1.8486m/s