A brick is thrown upward from the top of a building at an angle of 10° to the horizontal and with an initial speed of 11 m/s. If the brick is in flight for 2.8 s, how tall is the building?

Well, if the brick was thrown at an angle of 10°, we can assume it was auditioning for the role of a flying circus acrobat. Anyway, let's calculate the height of the building.

First, let's split the initial velocity of the brick into its horizontal and vertical components. The horizontal component is given by 11 m/s * cos(10°) and the vertical component is given by 11 m/s * sin(10°).

Now, let's focus on the vertical motion of the brick. We can use the equation: h = v₀t + (1/2)gt², where h is the height, v₀ is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.

We know that the initial vertical velocity is 11 m/s * sin(10°) and the time is 2.8 seconds. Since the brick is thrown vertically upward, the acceleration due to gravity is -9.8 m/s² (negative because it opposes the motion).

Using the equation, we find: h = (11 m/s * sin(10°)) * 2.8 s + (1/2) * (-9.8 m/s²) * (2.8 s)².

Now, let me grab my calculator and make this calculation... *beep boop beep*.

After crunching the numbers, it turns out the height of the building is approximately 16.69 meters.

So, the building is 16.69 meters tall. Just high enough to cause a few heart palpitations if you were to jump off it, but nothing too extreme.

To calculate the height of the building, we need to find the vertical component of the brick's initial velocity.

Step 1: Find the vertical component of the initial velocity.
The vertical component (Vy) can be found using the equation:
Vy = V * sin(θ), where V is the initial velocity and θ is the angle of projection.

Given:
Initial velocity (V) = 11 m/s
Angle of projection (θ) = 10°

Using the equation:
Vy = 11 * sin(10°)
Vy = 1.9 m/s (approx)

Step 2: Find the time taken for the brick to reach its maximum height.
When the brick reaches its maximum height, its vertical velocity becomes zero. The time taken (t) to reach the maximum height can be calculated using the equation:
t = Vy / g, where g is the acceleration due to gravity (9.8 m/s^2).

Given:
Vertical velocity (Vy) = 1.9 m/s
Acceleration due to gravity (g) = 9.8 m/s^2

Using the equation:
t = 1.9 / 9.8
t = 0.194 s (approx)

Step 3: Find the maximum height reached by the brick.
The maximum height (H) can be calculated using the equation:
H = Vy * t - (1/2) * g * t^2

Given:
Vertical velocity (Vy) = 1.9 m/s
Time taken to reach maximum height (t) = 0.194 s
Acceleration due to gravity (g) = 9.8 m/s^2

Using the equation:
H = 1.9 * 0.194 - (1/2) * 9.8 * (0.194^2)
H = 0.368 m (approx)

Step 4: Find the total height of the building.
The total height of the building is equal to the maximum height reached by the brick. Therefore, the height of the building is 0.368 m (approx).

To find the height of the building, we need to find the vertical displacement of the brick. We can do this by using the equation of motion for vertical motion:

y = v₀y * t + (1/2) * a * t²

Where:
y = vertical displacement (height of the building)
v₀y = initial vertical component of velocity
t = time in seconds
a = acceleration due to gravity (approximately -9.8 m/s²)

First, we need to determine the initial vertical component of velocity (v₀y). We can calculate this using the initial speed (11 m/s) and the angle of the projectile's path (10°).

v₀y = v₀ * sin(θ)

where:
v₀ = initial speed of the projectile
θ = angle of the projectile's path with respect to the horizontal

Plugging in the values:

v₀y = 11 m/s * sin(10°)

Now we can substitute this value into the equation for y:

y = (11 m/s * sin(10°)) * 2.8 s + (1/2) * (-9.8 m/s²) * (2.8 s)²

Simplifying the equation:

y = 11 m/s * sin(10°) * 2.8 s - 4.9 m/s² * 2.8 s²

Finally, calculate the value of y to find the height of the building.