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?

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

To find the height of the building, we need to analyze the vertical motion of the brick. We can break the motion into two components: one in the horizontal direction and one in the vertical direction.

First, let's determine the initial vertical velocity (v₀) of the brick.

Given:
- Initial speed (u) = 12 m/s
- Launch angle (θ) = 35°

We can use trigonometry to find the vertical component of the initial velocity:
v₀ = u * sin(θ)
v₀ = 12 * sin(35°)
v₀ ≈ 6.85 m/s

Now, we will use the following kinematic equation to find the height (h) of the building:

h = v₀ * t + (1/2) * g * t²

Given:
- Time (t) = 3.5 s (flight time)
- Acceleration due to gravity (g) = 9.8 m/s²

Substituting the values:
h = 6.85 * 3.5 + (1/2) * 9.8 * (3.5)²
h ≈ 23.98 + (1/2) * 9.8 * 12.25
h ≈ 23.98 + 14.98
h ≈ 38.96 meters

Therefore, the height of the building is approximately 38.96 meters.