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)
Vo = 15m/s[35o].
Yo = 15*sin35 = 8.60 m/s.
Y = Yo + g*Tr = 0, Tr = -Yo/g = -8.60/-9.8 = 0.88 s. = Rise time.
h1 = Yo*Tr + 0.5g*Tr^2.
h1 = 8.60*0.88 - 4.9*0.88^2 = 3.77 m. Above top of bldg.
Tr+Tf = 2.7, Tf = 2.7-Tr = 2.7-0.88 = 1.82 s. = Fall time from max ht. to gnd.
h2 = 0.5g*Tf^2 = 4.9*1.82^2 = 16.2 m. Above gnd.
Hb = h2-h1 = 16.2 - 3.77 = 12.5 m = Height of the bldg.
Tr+Tf = 2.7, Tf = 2.7-Tr =
Oops! Disregard last line of this post.
To determine the height of the building, we need to calculate the vertical displacement of the brick during its flight time.
First, let's break down the given information:
- Initial speed (u) = 15 m/s
- Launch angle (θ) = 35°
- Flight time (t) = 2.7 s
- Acceleration due to gravity (g) = 9.8 m/s² (assuming no air resistance)
Since the initial speed has both horizontal and vertical components, we can calculate the initial vertical velocity (uy) using the launch angle.
Vertical component of velocity (uy) = u * sin(θ)
uy = 15 m/s * sin(35°)
uy ≈ 8.60 m/s
Given that the brick is in flight for 2.7 seconds, we can calculate the vertical displacement (Δy) using the following equation:
Δy = uy * t - (1/2) * g * t^2
Substituting the known values:
Δy = 8.60 m/s * 2.7 s - (1/2) * 9.8 m/s² * (2.7 s)^2
Evaluating this equation gives us:
Δy ≈ 23.166 m
Therefore, the height of the building is approximately 23.166 meters.