The magnitude of the shell's acceleration just after it is released is 9.8 m/s²
Acceleration is the rate at which velocity is changing. Acceleration is a vector quantity and it is measured in m/s²
Given that a glaucous-winged gull, ascending straight upward at 5.00 m/s, drops a shell when it is 14.0 m above the ground. What is the magnitude of the shell's acceleration just after it is released?
Before glaucous-winged gull drops the shell, it has zero acceleration since it is moving up at a constant speed of 5m/s with the bird. When it lets go of the shell, the shell enters free fall. That means,
Acceleration a = Acceleration due to gravity g = 9.8 m/s²
Therefore, the magnitude of the shell's acceleration just after it is released is 9.8 m/s²
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The magnitude of the shell's acceleration just after it is released is equal to the acceleration due to gravity i.e 9.8 m/s²
This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
From the question given above, we were told that the shell started falling from a height of 14 m above the ground. At this stage, gravity is acting on the shell.
Thus, we can conclude that the magnitude of the shell's acceleration just after it is released is equivalent to the value of the acceleration due to gravity (i.e 9.8 m/s²).
NOTE: Gravity will keep acting on the shell until it reaches the ground.
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