A brick is thrown upward from the top of a building at an angle of 25° to the horizontal and with an initial speed of 17 m/s. If the brick is in flight for 2.5 s, how tall is the building?

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To find the height of the building, we need to break down the motion of the brick into its horizontal and vertical components.

First, let's find the initial vertical velocity of the brick. We can use the given initial speed and angle to calculate this.

Vertical Velocity (Vᵥ) = Initial Speed (v) * sin(angle)

Vᵥ = 17 m/s * sin(25°)
Vᵥ ≈ 17 m/s * 0.4226
Vᵥ ≈ 7.19 m/s

Next, let's find the time it takes for the brick to reach its maximum height. Since the motion is symmetric, the time taken to reach the maximum height will be half the total time of flight.

Time to reach maximum height (t) = Total Time of Flight (T) / 2
t = 2.5 s / 2
t = 1.25 s

Now, we can find the height reached by the brick using the formula for vertical displacement:

Vertical Displacement (h) = Initial Vertical Velocity (Vᵥ) * Time of Flight (T) - 0.5 * gravitational acceleration (g) * Time of Flight (T)^2

where gravitational acceleration (g) is approximately 9.8 m/s².

h = Vᵥ * t - 0.5 * g * t^2
h = 7.19 m/s * 1.25 s - 0.5 * 9.8 m/s² * (1.25 s)^2
h = 8.987 m - 7.225 m
h ≈ 1.76 m

Therefore, the height of the building is approximately 1.76 meters.