A cannonball is shot horizontally off a high castle wall at 47.4 m/s. What is the magnitude of the cannonball's velocity after 1
1 answer:
The magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.
<h3>What is velocity?</h3>The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
u is the initial velocity of canonball= 47.4 m/sec
g is the acceleration of free fall = 9.81 m/sec²
v is the velocity after 1.23 s
According to Newton's first equation of motion,
Hence, the magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.
To learn more about the velocity, refer to the link: brainly.com/question/862972.
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Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
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