A rock is shot vertically upward from the edge of the top of a tall building.The rock reaches its maximum height above the top of the building 1.90 s after being shot. Then, after barely missing the edge of the building as it falls downward, the rock strikes the ground 9.00 s after it is launched. In SI units: (a) with what upward velocity is the rock shot, (b) what maximum height above the top of the building is reached by the rock, and (c) how tall is the building?

Answer 1

(a)
Upward motion

v=v₀ -gt
v=0
v₀ =gt =9.8•1.9 =18.62 m/s
(b)
h=v₀t -gt²/2 = 18.62•1.9 - 9.8•1.9²/2=
=35.38 – 17.69 = 17.69 m
(c)
If H is the height of the building
t₀= 9 – 1.9 = 7.1 s
H+h = gt₀²/2 = 9.8•7.1²/2 =247 m
H=247 –h= 247-17.69 =229.5 m