A brick is thrown upward from the top of a building at an angle of 35° to the horizontal and with an initial speed of 12 m/s. If the brick is in flight for 3.5 s, how tall is the building?

Answer 1

Vi = 12 sin 35 = 6.88 m/s

how far above building does it go?
V top = 0 = 6.88 - 9.8 t
t = .7 seconds up
ke = 0 at top = (1/2) v^2 - gh
.5(6.88)^2 = 9.8 h
h = 2.42 meters above building

now, new problem. drop brick from 2.42 meters above building
time falling = 3.5 - .7 = 2.8 seconds
h = 4.9 (2.8)^2
= 38.4 from top
so building is 38.4 - 2.4 = 36 meters high