An arrow fired from the bow vertically upward at a speed of 30 m / s hits the target after 2

An arrow fired from the bow vertically upward at a speed of 30 m / s hits the target after 2 seconds. What speed did the arrow have at the time of reaching the target, and at what height was the target?

Given:

v = 30 m / s (meters per second) – the initial speed of the arrow released vertically upwards;

t = 2 seconds – the time interval after which the arrow hits the target.

It is required to determine v1 (m / s) – the speed of the arrow at the time of reaching the target and h (meter) – the height at which the target was.

The arrow speed when hitting the target will be equal to:

v1 = v – g * t, where g = 9.8 Newton / kilogram;

v1 = 30 – 2 * 9.8 = 30 – 19.6 = 10.4 m / s.

Then the height of the target will be equal to:

h = v * t – g * t ^ 2/2 = 30 * 2 – 9.8 * 2 ^ 2/2 = 60 – 9.8 * 2 = 60 – 19.6 = 40.4 meters above the ground.

Answer: the speed of the arrow upon reaching the target is 10.4 m / s, the target was at a height of 40.4 meters above the ground.