One man is on a 60 m high building, releasing a rock. After 2 s, another man on the ground throws the ball upright at a speed of 10 m / s. Find that both objects will stand high apart from the ground. how much

The distance apart from the rock to the ball just before the ball is thrown is taken to be approximately the distance of the rock to the ground after 2 seconds

The height of the rock from the ball from the ground after 2 seconds is approximately 40.38 meters

Explanation:

The given parameters are;

1) The height of the building fro which the first man releases the rock = 60 m

2) The speed with which the other man throws the ball vertically upwards = 10 m/s

The time after the release of the rock that the ball was thrown up = 2 seconds

The speed, v, of the rock after the 2 seconds = u  + gt

v = 0 + 9.81 × 2 = 19.62 m/s

The position of the rock, s, after 2 seconds = (v² - u²)/(2×g)

s = (19.62² - 0²)/(2×9.81) = 19.62 meters

The distance from the ground = Height of building - s = 60 - 19.62 = 40.38 m

Therefore, bust before the ball is thrown, the distance between the ball and the rock ≈ 40.38 meters.

f=ma newton

f=100x2.5=250n

not too sure thinking all accelerations are pointed down