A body weighing 1 kg freely falls from a certain height. At the moment of falling to the ground, its kinetic energy is 98 J.

A body weighing 1 kg freely falls from a certain height. At the moment of falling to the ground, its kinetic energy is 98 J. From what height does the body fall? And what is the speed of the body the moment it hits the ground.

If the mass is m = 1 kg, and the kinetic energy at the moment of falling Ek = 98 J, then we will use the formula:

Ek = m * V ^ 2/2, where V is the final velocity of the object.

Means:

V2 = 2 * Ek / m;

V = √‾ (2 * Ek / m);

V = √‾ (2 * Ek / m) = √‾ (2 * 98/1) = √‾196 = 14 m / s;

In this case, V = V0 + a * t, where a is the acceleration of the body, t is the time of movement, and V0 is the initial velocity. The acceleration in the case of a fall is equal to the acceleration in free fall – g, and V0 = 0, therefore:

V = g * t.

Substitute in the formula:

Ek = m * V ^ 2/2 = m * g ^ 2 * t ^ 2/2;

The fall time is unknown to us, but for t2 we can derive the formula:

t 2 = 2 * Ek / (m * g ^ 2);

Formula describing the height from which the body fell:

h = g * t ^ 2/2 = g * 2 * Ek / (2 * m * g ^ 2) = Ek / (m * g);

h = Ek / (m * g) ≈ 98 / (1 * 9.8) = 10 m.

Answer: the body fell from a height of h = 10 m. The final velocity of the body was V = 14 m / s.