The body is thrown vertically upward with an initial speed of 31 m / s.

The body is thrown vertically upward with an initial speed of 31 m / s. Find the path traversed by the body in the fourth second after the start of the movement.

Let us express the original distance using the formula for moving a body thrown vertically upward:

h = v0 * t – g * t ^ 2/2.

Where h is the desired distance, v0 is the initial speed, according to the condition 31 m / s, g is the acceleration of gravity, a constant value of 10 m / s ^ 2, t is the time, according to the condition 4 s.

Let’s find the falling speed:

V = vо – gt = 31 – 10 * 4 = 31 – 40 = -9 m / s.

A negative speed indicates that the body was raised to a height, stopped and began to fall to a height:

H = vo * t – gt ^ 2/2 = 31 * 4 – 10 * 16/2 = 124 – 80 = 44 meters.

Let’s calculate how long it went up:

0 = 31 – 10 * t1;

t1 = 3.1 s went up and 0.9 s went down.

It remains to find to what height it was raised in 3.1 s:

H = 31 * 4 – 10 * (3.1) ^ 2/2 = 124 – 96.1 / 2 = 75.95 meters.

75.95 – 44 = 31.95 meters.