A brick is thrown upward from the top of a building at an angle of 35° to the horizontal and with an initial speed of 15 m/s. If the brick is in flight for 2.7 s, how tall is the building? in (m)
y=1/2gt^2+vsin (35)
y=1/2×10×2.7^2+15×sin (35)
y=30.02m
See previous post: Thu, 2-4-16, 10:47 PM.
To find the height of the building, we need to analyze the motion of the brick. Let's break down the problem into horizontal and vertical components.
1. Horizontal Motion:
The horizontal component of the initial velocity remains unchanged throughout the entire motion. Since there is no acceleration in the horizontal direction, the horizontal velocity (Vx) is constant.
Vx = initial velocity * cos(angle)
Vx = 15 m/s * cos(35°)
Vx = 15 m/s * 0.819
Vx ≈ 12.28 m/s
2. Vertical Motion:
The brick is thrown upward and experiences a constant acceleration due to gravity acting in the downward direction.
We can use the equation for vertical motion to find the initial vertical velocity (Vy) and the height (h):
Vy = initial velocity * sin(angle)
Vy = 15 m/s * sin(35°)
Vy = 15 m/s * 0.574
Vy ≈ 8.61 m/s
Using the equation for vertical motion:
h = Vy * t + (1/2) * g * t^2
h = 8.61 m/s * 2.7 s + (1/2) * 9.8 m/s^2 * (2.7 s)^2
h = 23.25 m + 35.63 m
h ≈ 58.88 m
Therefore, the height of the building is approximately 58.88 meters.