# What is the path of a freely falling body in 3 seconds? The initial velocity is zero.

**What is the path of a freely falling body in 3 seconds? The initial velocity is zero. What speed will the body have after the specified time?**

t = 3 s.

g = 10 m / s2.

V0 = 0 m / s.

S -?

V -?

On a freely falling body, only gravity acts m * g, so it moves with the acceleration of gravity g.

For uniformly accelerated motion of a body, its path S is determined by the formula: S = V0 * t + g * t ^ 2/2, where V0 is the initial velocity of the body, t is the time of motion of the body.

Since the body falls V0 = 0 m / s, then S = g * t ^ 2/2.

S = 10 m / s2 * (3 s) ^ 2/2 = 45 m.

The speed of the body V with uniformly accelerated motion is expressed by the formula: V = V0 + g * t = g * t.

V = 10 m / s2 * 3 s = 30 m / s.

Answer: the body will travel a path S = 45 m and will have a speed V = 30 m / s.