A stone was thrown from a height of 2 m at an angle to the horizon with an initial speed of 6 m / s.

A stone was thrown from a height of 2 m at an angle to the horizon with an initial speed of 6 m / s. Find the speed of the stone as it falls to the ground.

Eк1 + Eп1 = Eк2 + Eп2, Eк1 and Eп1 – energy when throwing; Eк2 and Ep2 – energy at the moment of falling.

When falling to the ground, h = 0 m and Ep2 = 0.

Eк1 + Ep1 = Eк2.

Eк1 = m * V0² / 2, m is the body mass, V0 is the initial velocity of the stone (V = 6 m / s).

Ep1 = m * g * h, g is the acceleration of gravity (g = 10 m / s²), h is the height from which the stone is thrown (h = 2 m).

Eк2 = m * V² / 2, V0 – speed at the moment of falling.

m * V0² / 2 + m * g * h = m * V² / 2.

V0² / 2 + g * h = V² / 2.

V = sqrt (V0² + 2g * h) = 8.72 m / s.