A body weighing 0.1 kg is thrown horizontally at a speed of 4 m / s from a height of 2 m relative to the earth’s surface.

A body weighing 0.1 kg is thrown horizontally at a speed of 4 m / s from a height of 2 m relative to the earth’s surface. What is the kinetic energy of the body at the moment of its landing? Disregard air resistance.

Data: m (thrown body weight) = 0.1 kg; Vg (horizontal speed) = 4 m / s; h is the height of the body throw (h = 2 m).

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

1) Vertical speed at the moment of landing: m * Vv ^ 2/2 = m * g * h, whence we express: Vv = √ (2 * g * h) = √ (2 * 10 * 2) = 6.32 m / with.

2) Full speed at the moment of landing: V = √ (Vv ^ 2 + Vg ^ 2) = √ (6.32 ^ 2 + 4 ^ 2) = 7.48 m / s.

3) Kinetic energy: Ek = m * V ^ 2/2 = 0.1 * 7.48 ^ 2/2 = 2.8 J.

Answer: The kinetic energy was equal to 2.8 J.