The ball rolls on a smooth horizontal surface of the table at a constant speed of 2 m / s.

The ball rolls on a smooth horizontal surface of the table at a constant speed of 2 m / s. Having reached the edge of the table, the ball falls to the floor at a distance of 0.8 m from the base of the table. Find the height of the table.

Vg = 2 m / s.

S = 0.8 m.

g = 10 m / s2.

h -?

The motion of the ball can be divided into two types: horizontally, it moves uniformly at a speed Vg, and vertically moves with an acceleration of free fall g from a state of rest.

S = Vg * t, where t is the time the ball falls.

Let’s find the time of movement of the ball t: t = S / Vg.

t = 0.8 m / 2 m / s = 0.4 s.

Let us express the height of the fall of the ball h by the formula: h = g * t ^ 2/2.

h = 10 m / s2 * (0.4 s) ^ 2/2 = 0.8 m.

Answer: the height of the table from which the ball falls is h = 0.8 m.