From a height of 1.25 m, a ball was thrown vertically downward at a speed of 10 m / s.

From a height of 1.25 m, a ball was thrown vertically downward at a speed of 10 m / s. To what height after impact will it rise if 40 percent of the ball’s mechanical energy is lost during the impact?

h1 = 1.25 m.

V1 = 10 m / s.

Q = k * E1 / 100%.

g = 10 m / s2.

k = 40%.

h2 -?

The total mechanical energy of the body E is the sum of the potential energy En and the kinetic energy of the body Ek: E = En + Ek.

At the beginning of the throw, the total mechanical energy E1 consists of the potential energy En1 and the kinetic energy Ek1: E1 = En1 + Ek1.

The potential energy of the ball En1 is determined by the formula: En1 = m * g * h1, where m is the mass of the ball, g is the acceleration of gravity, h1 is the height from which the ball falls.

The kinetic energy of the ball Ek1 is determined by the formula: Ek1 = m * V1 ^ 2/2, where m is the mass of the ball, V is the speed of the ball when thrown.

When bouncing to the maximum height h2, the ball stops, the total mechanical energy E2 consists only of potential En2: E2 = En2.

En2 = m * g * h2.

E1 = E2 + Q = E2 + k * E1 / 100%.

m * g * h1 + m * V1 ^ 2/2 = m * g * h2 + k * (m * g * h1 + m * V1 ^ 2/2) / 100%.

g * h1 + V1 ^ 2/2 = g * h2 + k * (g * h1 + V1 ^ 2/2) / 100%.

h2 = h1 + V1 ^ 2/2 * g – k * (h1 + V1 ^ 2/2 * g) / 100%.

h2 = 1.25 m + (10 m / s) 2/2 * 10 m / s2 – 40% * (1.25 m + (10 m / s) 2/2 * 10 m / s2) / 100% = 3.75 m.

Answer: the ball rises to a height of h2 = 3.75 m.