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(m)/(s)cos(61\º)=4.169(m)/(s)` (2)

`V_(oy)=V_(o)sin\alpha` (3)

`V_(oy)=8.6(m)/(s)sin(61\º)=7.521(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= (X)/( V_(ox))` (6)

`t=2.42s` (7)

In the y-axis:

`-y=V_(oy).t+(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(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(m)/(s))(2.42s)+(1)/(2)(-9.8(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