The body from a state of rest freely falls to the ground for 4 seconds. From what height it falls

The body from a state of rest freely falls to the ground for 4 seconds. From what height it falls at what speed it will have at the time of landing.

Body drop height:

S = V0 * t + a * t² / 2, where V0 is the initial velocity of the body (the body fell freely, V0 = 0 m / s), and is the acceleration of the body (a = g, g is the acceleration of gravity (we take g = 10 m / s²)), t – fall time (t = 4 s).

S = g * t² / 2 = 10 * 42/2 = 80 m.

Body speed at the moment of landing:

S = (V² – V0²) / 2a.

S = V² / 2g.

V = √ (2g * S) = √ (2 * 10 * 80) = √ (1600) = 40 m / s.

Answer: The body fell from a height of 80 m and had a speed of 40 m / s at the time of landing.