# A steel ball weighing 50 g was thrown vertically upward with an initial velocity of 20 m / s.

**A steel ball weighing 50 g was thrown vertically upward with an initial velocity of 20 m / s. Determine the potential and kinetic energy of the ball 1 second after the start of movement. The value of the potential energy of the ball at the initial moment of time is taken equal to zero.**

m = 50 g = 0.05 kg.

g = 10 m / s2.

V0 = 20 m / s.

t = 1 s.

Ek -?

Ep -?

The potential energy En and the kinetic energy Ek of the body are determined by the formulas: En = m * g * h, Ek = m * V ^ 2/2.

The body moves upward with the acceleration of gravity g, so the height of rise h is found by the formula: h = V0 * t – g * t2 / 2.

h = 20 m / s * 1 s – 10 m / s2 * (1 s) ^ 2/2 = 15 m.

Ep = 0.05 kg * 10 m / s * 15 m = 7.5 J.

We find the kinetic energy of the body Ek from the law of conservation of total mechanical energy: Ek = Ek0 – En.

Ek0 = m * V0 ^ 2/2.

Ek0 = 0.05 kg * (20 m / s) ^ 2/2 = 10 J.

Ek = 10 J – 7.5 J = 2.5 J.

Answer: En = 7.5 J, Ek = 2.5 J.