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

Answer:

38.47 m

Explanation:

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

$y_f=y_i+v_{iy}t+\frac{1}{2}at^2$

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

$0=y_i+2.78(3.1)+\frac{1}{2}(-9.8)(3.1)^2$

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