The apple fell from a height of h = 2.2 m during t = 0.67 s. Was the falling speed of the apple constant?

| January 3, 2021 | education

The apple fell from a height of h = 2.2 m during t = 0.67 s. Was the falling speed of the apple constant? What is its average value?

h = 2.2 m.
t = 0.67 s.
Vav -?
When the apple falls, the force of gravity, constant in value and direction, acts, therefore, according to 2 Newton’s laws, the apple moves uniformly with the acceleration of gravity g = 9.8 m / s2.
This means that for every second of movement t = 1 s, the speed of the apple increases by ΔV = 9.8 m / s.
V = g * t.
V = 9.8 m / s2 * 0.67 s = 6.57 m / s.
To find the average speed of movement Vav, it is necessary to divide the entire path h passed by the apple by the time of its movement t: Vav = h / t.
Vav = 2.2 m / 0.67 s = 3.28 m / s.
Answer: the speed of the apple increases with the acceleration of gravity from 0 to V = 6.57 m / s, the average speed of the apple is Vav = 3.28 m / s.



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