stone is dropped from building at same time another stone is projected vertically upwards at 20m/s

Question

The stones pass each other after 3 seconds. What is the height of the building

Answer 1

Let Y₁(t) be position of dropped stone and Y₂(t) be position of projected stone.

Then:

\(​Y_1(t) = H - \frac12gt^2\)

\(Y_2(t) = v_ot - \frac12gt^2 ​\)

\(​ Y_1(3) = Y_2(3)\implies H - \frac92g = 20 \frac ms\cdot 3s - \frac92g ​\)

Therefore \(H = 60m\)