Question

A stone is dropped from the roof of a high building. A second stone is dropped 2.00 s later. What is the position of the second stone relative to the first stone after 4.00 seconds?

Answer 1

After 4 seconds, the second stone that was dropped 2 seconds after the first will be 58.8 m below the first stone.

Step-by-step explanation:

When the first stone is dropped from a building, it starts with an initial velocity of zero and accelerates downward due to gravity. Assuming acceleration due to gravity is 9.8 m/s², after 4.00 seconds, it would have fallen a distance given by the equation s = ut + ½at², where u is the initial velocity, a is the acceleration, and t is the time. Substituting the values, we get s = 0 + ½(9.8)(4.00²) = 78.4 m. However, the second stone is dropped 2.00 seconds after the first and so has only been falling for 2.00 seconds by the time 4.00 seconds have elapsed since the first was dropped. Thus, for the second stone, t = 2.00 s and the distance it falls is s = 0 + ½(9.8)(2.00²) = 19.6 m. Therefore, the position of the second stone relative to the first after 4.00 seconds is 78.4 m - 19.6 m = 58.8 m below the first stone.