# A coin is dropped from a height of 90 meter. How fast will it fall, just before it hits the ground?

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### 1 Answer

This is a problem that can be solved using the kinematic method, through the equations for vertical movement; or by the method of energy, using the law of conservation of mechanical energy. Considering the proposed suggestion, we will apply the second option.

According to the law of conservation of mechanical energy, the sum of the kinetic energy and gravitational potential energy, remains constant in every position during the fall of the coin. Let's call the point of maximum height as position 1 and the point, just before touching the ground, as position 2; then we can write the following equation:

Ek(1) + Ep(1) = Ek(2) + Ep(2)

(m*v1^2)/2 + (m*g*h1) = (m*v2^2)/2 + ( m*g*h2) (1)

m, is the mass of the coin.

v, is the velocity at any position.

h, is the height in any position.

g, is the acceleration of gravity.

In our case we have the following values:

v1 = 0, in the highest position.

h2 = 0, just before touching the ground.

Substituting in (1):

0 + (m*g*h1) = (m*v2^2)/2 + 0

v2^2 = 2*g*h1

v2 = sqrt (2*g*h1) = sqrt [2(10)(90)]

**v2 = 42.42 m/s**