Question

A brick is thrown upward from the top of a building at an angle of 10° to the horizontal and with an initial speed of 16 m/s. If the brick is in flight for 3.1 s, how tall is the building?_____ m

Answer 1

38.47 m

Step-by-step explanation:

To find the height of the building, we will use the following equation

`y_f=y_i+v_(iy)t+(1)/(2)at^2`

Where yf is the final height, yi is the initial height, viy is the initial vertical velocity, t is the time, and a is the acceleration due to gravity.

If the brick is in flight for 3.1 s, we can say that when t = 3.1s, yf = 0 m. So, replacing

viy = (16 m/s)sin(10) = 2.78 m/s

a = -9.8 m/s²

we get

`0=y_i+2.78(3.1)+(1)/(2)(-9.8)(3.1)^2`

Solving for yi

`\begin{gathered} 0=y_i-38.48 \\ y_i=38.48\text{ m} \end{gathered}`

Therefore, the height of the building is 38.48 m