A brick is thrown upward from the top of a

building at an angle of 25.1

above the horizontal and with an initial speed of 19.4 m/s.
The acceleration of gravity is 9.8 m/s
2
.
If the brick is in flight for 3.5 s, how tall is

To determine how tall the building is, we need to find the height reached by the brick during its flight.

To solve this problem, we can break down the motion of the brick into horizontal and vertical components.

1. Horizontal Motion:
The question states that the brick is thrown at an angle of 25.1 degrees above the horizontal. Since there is no horizontal acceleration, we can assume that the horizontal velocity remains constant throughout the motion. Therefore, the horizontal component of the velocity is given by:

Vx = V * cos(θ)
= 19.4 m/s * cos(25.1°)
≈ 17.646 m/s

This horizontal velocity remains constant during the entire flight.

2. Vertical Motion:
The vertical component of the velocity changes due to the acceleration of gravity. We can use the kinematic equation:

Δy = Vyi * t + (1/2) * g * t^2

where:
Δy is the displacement in the vertical direction (height of the building),
Vyi is the initial vertical velocity, which is given by:

Vyi = V * sin(θ)
= 19.4 m/s * sin(25.1°)
≈ 8.343 m/s

Substituting the given values into the equation:

Δy = (8.343 m/s) * (3.5 s) + (1/2) * (9.8 m/s^2) * (3.5 s)^2
≈ 72.309 m

Therefore, the building is approximately 72.309 meters tall.