First, we have to establish the <u>reference system</u>. 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:

$X=V_{ox}.t$ (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)

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

Where:

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

$g=-9.8\frac{m}{s^{2}}$ 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

3 0

3 years ago

earnstyle [38]

Answer:

0.0312J

Explanation:

Let x be the distance the staple moves:

$x=0.150m-0.115m=0.035m$

And spring constant is $k=51.0N/m$

$PE=0.5kx^2\\=0.5\times 51.0\times 0.035^2\\\\=0.312$

Hence, the potential energy is 0.0312J

8 0

3 years ago