A ball is rolling horizontally at 3.00 meters per second as it leaves the edge of a tabletop 0.750 meter above the floor. The ball lands on the floor 0.391 second after leaving the tabletop. What is the magnitude of the ball’s acceleration 0.200 second after it leaves the tabletop? [Neglect friction.]

This numerical value for the acceleration of a free-falling object is given a special name "the acceleration of gravity" - the acceleration for any object moving under the sole influence of gravity. The value for the acceleration of gravity is most accurately known as [tex]9.8 m/s^{2}[/tex]. There are slight variations in this numerical value (to the second decimal place) that are dependent on altitude.

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-11-05T18:57:12+01:00

The magnitude of the ball’s acceleration 0.200 second after it leaves the tabletop is 9.81 m/s².

The given parameters;

horizontal speed of the ball, u = 3 m/s
height of the table, h = 0.75 m
time o motion of the ball, t = 0.391 s

An object falling downwards is under the influence of acceleration of due to gravity. The acceleration of due to gravity is always directed downwards and it is constant.

[tex]h = v_0_yt + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\2h = gt^2\\\\g = \frac{2\times h}{t^2} \\\\g = \frac{2 \times 0.75}{0.391^2} \\\\g = 9.81 \ m/s^2[/tex]

Thus, the magnitude of the ball’s acceleration 0.200 second after it leaves the tabletop is 9.81 m/s².

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