A brick is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 11 m/s. If the brick is in flight for 3.2 s, how tall is the building?

___m

To find the height of the building, we need to determine the maximum height reached by the brick during its flight time.

Step 1: Resolve the initial velocity into horizontal and vertical components:

The initial speed of the brick, 11 m/s, can be divided into its horizontal and vertical components.
The horizontal component can be calculated as:
Vx = V * cos(theta)
= 11 m/s * cos(30°)
= 11 m/s * sqrt(3)/2
= 9.5 m/s (approximately)

The vertical component can be calculated as:
Vy = V * sin(theta)
= 11 m/s * sin(30°)
= 11 m/s * 1/2
= 5.5 m/s

Step 2: After the brick is thrown upward, it will follow a projectile motion.
The equation for the height of a projectile at any time (t) can be given as:
h = Vy * t - (1/2) * g * t^2
where,
h = height
Vy = vertical component of initial velocity
t = time
g = acceleration due to gravity (approximately 9.8 m/s^2)

Step 3: Substitute the values into the equation:

h = 5.5 m/s * 3.2 s - (1/2) * 9.8 m/s^2 * (3.2 s)^2
h = 17.6 m - 50.176 m
h ≈ -32.576 m

Since the height cannot be negative, we ignore the negative sign and take the absolute value.

Final Answer:
The height of the building is approximately 32.58 meters.

To find the height of the building, we can use the following kinematic equation:

h = h0 + v0*t + (1/2)*a*t^2

Where:
h = height of the building
h0 = initial height of the brick
v0 = initial velocity of the brick
t = time of flight
a = acceleration

In this case, the brick is thrown upward, so the acceleration is equal to the acceleration due to gravity, which is approximately -9.8 m/s^2 (negative because it is acting in the opposite direction of the motion).

First, let's calculate the initial vertical velocity (v0y) of the brick:
v0y = v0 * sin(theta)

Where:
v0y = initial vertical velocity
v0 = initial velocity of the brick
theta = angle of projection

v0y = 11 m/s * sin(30°)
v0y = 11 m/s * 0.5
v0y = 5.5 m/s

Next, we can find the vertical position (h) of the brick after 3.2 seconds using the kinematic equation:

h = h0 + v0y*t + (1/2)*a*t^2

Since the brick is thrown from the top of the building, the initial height (h0) is equal to the height of the building.

Substituting the given values:
h = h0 + (5.5 m/s) * (3.2 s) + (1/2) * (-9.8 m/s^2) * (3.2 s)^2

Let's solve this equation to find the height of the building.