Question

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.]

Answer 1

Answer: 9.81m/s^2

Explanation: since it’s free fall and friction is neglected, the magnitude of the ball’s acceleration will be equal to the general acceleration due to gravity which is 9.81m/s^2

Answer 2

Answer:

$9.8 m/s^{2}$

Explanation:

Any object in free-fall will accelerate due to gravity. Acceleration due to gravity is equal to $9.8 m/s^{2}$.

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 $9.8 m/s^{2}$. There are slight variations in this numerical value (to the second decimal place) that are dependent on altitude.