Two pebbles were released from the hands from the same height, one after the other after 1 s.

Two pebbles were released from the hands from the same height, one after the other after 1 s. By what law will the distance between them change as they fall further?

According to the condition of the problem, the second pebble began to fall 1 second after the first ball.

The formula for calculating the distance that the first ball has passed will look like:

h1 = g * t ^ 2/2 where g is the acceleration of gravity

since the second balloon began its flight with a difference of 1 second, the formula for the second balloon will look like this:

h2 = g * (t – 1) ^ 2/2

The distance between the balls will be:

S = h1 – h2 = g * t ^ 2/2 – g * (t – 1) ^ 2/2 = g / 2 * (t ^ 2 – (t-1) ^ 2) = g / 2 * (t ^ 2 – t ^ 2 + 2 * t – 1 ^ 2) = g / 2 * (2 * t – 1).

Taking g = 10 m / s ^ 2, we get:

S = g / 2 * (2 * t – 1) 10/2 * (2 * t – 1) = 5 * (2 * t -1).

Answer: the distance between them will change according to the law S = 5 * (2 * t -1).