A brick is thrown upward from the top of a building at an angle of 40° to the horizontal and with an initial speed of 16 m/s. If the brick is in flight for 3.4 s, how tall is the building?
To find the height of the building, we can use the kinematic equations of motion. We need to consider the vertical motion of the brick since we want to find the height.
Let's break down the information given:
- Initial velocity (u) = 16 m/s (upward)
- Angle of projection (θ) = 40°
- Time of flight (t) = 3.4 s
First, we need to find the vertical component of the initial velocity (v₀y) using trigonometry. The vertical component can be calculated as:
v₀y = u * sin(θ)
v₀y = 16 m/s * sin(40°)
v₀y = 10.275 m/s (upward)
Next, we need to use the time of flight to find the total vertical displacement (h). The formula for vertical displacement is:
h = v₀y * t - (1/2) * g * t²
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Now, we substitute the values into the equation:
h = (10.275 m/s) * (3.4 s) - (1/2) * (9.8 m/s²) * (3.4 s)²
h = 34.965 m - 55.604 m
h = -20.64 m
Since the displacement is negative, it means the brick has fallen below its initial height. Therefore, the height of the building is 20.64 meters.
At time t after throwing,
Y (measured from the ground) = H + 16sin40*t - 4.9 t^2
H is the building height.
Set Y = 0 at t = 3.4s and solve for H
H = 0 + 4.9*(3.6)^2 - 34.97 = 28.53 m