Answer:
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
Step-by-step 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.