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 3.3 s, how tall is the building?

m

the height of a freely falling object is: H = .5g*t^2 + Vo*t + Ho

Vo is the initial vertical velocity

Ho is the initial height (height of building)

the height is 0 when the brick hits the ground in 3.3 sec

g is -9.8 m/s^2

H = .5gt^2 + [17 * sin(25º)]t + Ho

0 = -4.9(3.3^2) + [17 * sin(25º)]3.3 + Ho

0 = -53.4m + 23.7m + Ho

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

First, let's find the initial velocity of the brick in the vertical direction (Vy) and the horizontal direction (Vx).

Vy = V * sinθ
Vx = V * cosθ

where V is the initial speed of the brick (17 m/s) and θ is the angle of projection (25°).

Vy = 17 m/s * sin(25°)
Vy ≈ 7.12 m/s

Vx = 17 m/s * cos(25°)
Vx ≈ 15.26 m/s

Next, we can calculate the time it takes for the brick to reach its maximum height. Since the brick is thrown vertically, the initial vertical velocity (Vy) will decrease until it reaches zero at the maximum height, and then it will start to fall. Therefore, the time taken to reach the maximum height will be half of the total flight time.

t_max = t_flight / 2
t_max = 3.3 s / 2
t_max = 1.65 s

Using this time, we can find the maximum height (H_max) reached by the brick vertically using the equation of motion:

H_max = Vy * t_max - (1/2) * g * t_max^2

where g is the acceleration due to gravity (9.8 m/s^2).

H_max = 7.12 m/s * 1.65 s - (1/2) * 9.8 m/s^2 * (1.65 s)^2
H_max ≈ 5.97 m

Finally, to find the height of the building, we need to account for the initial height from where the brick was thrown. If the brick was thrown from the top of the building, the height of the building would be equal to the maximum height reached by the brick.

Therefore, the height of the building is approximately 5.97 meters.