a stone is drop from the top of a building and at thesame time a second stone is thrown vertically upward from the botom of the building at a speed of 20m/s, they passed each other 3s later. find the hight of the building.

Answer 1

Newton's law of motion wrt to displacement:

d=v_i * t + 0.5*a*t^2

For the stone being dropped from building:

a=accerlation=gravity=9.81m/s^2

d=0+0.5*(9.81)*3^2

For the second stone being thrown up vertically:

d=20*3+0.5*(-9.81)*3^2

Since displacement/height is same at 3 seconds:

d-0.5*(9.81)*3^2 = 20*3+0.5*(-9.81)*3^2

Solve for d

Answer 2

so the increase in speed of the dropped one cancels the slow down of the other and
height = initial speed up * time
= 20*3 = 60