A ball was thrown vertically upwards from a balcony located at a height of 4 m from the ground at a speed of 9 m / s.

A ball was thrown vertically upwards from a balcony located at a height of 4 m from the ground at a speed of 9 m / s. Where is the ball after a time of 2 seconds from the moment of throwing it relative to the Earth?

Given:

H0 = 4 meters – the height of the balcony above the ground;

v0 = 9 m / s – the speed with which the ball was thrown vertically upward;

t = 2 seconds – the time the ball moves;

g = 10 m / s ^ 2 – acceleration of gravity.

It is required to determine H (meters) – the position of the ball above the ground in time t.

Let’s compose the equation of the ball movement along the Y-axis, where 0 is the surface of the earth. Then:

H = H0 + v0 * t – g * t ^ 2/2

H = 4 + 9 * 2 – 10 * 2 ^ 2/2 = 4 + 18 – 10 * 4/2 = 22 – 20 = 2 meters.

Answer: After 2 seconds, the ball will be 2 meters above the ground.