Kinematics Question

Question

Two stones are released simultaneously: one at the top of a building and the other at the base of the building. The height of the building is 21 metres. The stone at the top of the building is released from rest. With what speed must the stone from the bottom of the building be thrown upward so that the two stones collide at a point 14 metres above the base of the building?

Answer 1

The stone at the top of the building must fall 7 feet.

21 - 4.9t^2 = 7

t^2 = 140/49

t = (2/7)sqrt(35)

In that amount of time the stone from the bottom will travel: vt - 4.9t^2 = 14 (where t is the value above)

and v(2sqrt(35)/7) = 14 + 4.9(140/49) = 28

and v = (14/5)sqrt(35)