Question

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how tall is the building?

Answer 1

Answer: 10.52m

First, we have to establish the reference system. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$:

$V_{ox}=V_{o}cos\alpha$   (1)

$V_{ox}=8.6\frac{m}{s}cos(61\º)=4.169\frac{m}{s}$  (2)

$V_{oy}=V_{o}sin\alpha$   (3)

$V_{oy}=8.6\frac{m}{s}sin(61\º)=7.521\frac{m}{s}$   (4)

As this is a projectile motion, we have two principal equations related:

In the x-axis:

$X=V_{ox}.t$  (5)

Where:

$X=10.1m$ is the distance where the brick landed

$t$ is the time in seconds

If we already know $X$ and $V_{ox}$, we have to find the time (we will need it for the following equation):

$t= \frac{X}{ V_{ox}}$  (6)

$t=2.42s$  (7)

In the y-axis:

$-y=V_{oy}.t+\frac{1}{2}g.t^{2}$   (8)

Where:

$y$ is the height of the building (in this case it has a negative sign because of the reference system we chose)

$g=-9.8\frac{m}{s^{2}}$ is the acceleration due gravity

Substituting the known values, including the time we found on equation (7) in equation (8), we will find the height of the building:

$-y=(7.521\frac{m}{s})(2.42s)+\frac{1}{2}(-9.8\frac{m}{s^{2}}).(2.42s)^{2}$   (9)

$-y=-10.52m$   (10)

Multiplying by -1 each side of the equation:

$y=10.52m$ >>>>This is the height of the building