Question

A rock is shot up vertically upward from the edge of the top of the building. The rock reaches its maximum height 2 s after being shot. Them, after barely missing the edge of the building as it falls downward, the rock strikes the ground 8 s after it was launched. Find
(a) upward velocity the rock was shot at;
(b) the maximum height above the building the rock reaches; and
(c) how tall is the building?

Answer 1

a. V = Vo + g*t = 0.
Vo - 9.8*2 = 0.
Vo = 19.6 m/s.

b. h = Vo*t + 0.5g*t^2.
g = -9.8 m/s^2.
t = 2 s.
h = ?.

c. Tr+Tf = 2 + 2 = 4 s. To fall back to
top of bldg. Tr = Rise time. Tf = Fall
time.

8-4 = 4 s. To fall from top of bldg. to
gnd.

h = Vo*t + o.5g*t^2.
Vo = 19.6 m/s.
t = 4 s.
g = 9.8 m/s^2.
h = ?