# Question fc978

Jul 6, 2015

The height of the building: 45 m.
The angle of contact: 63.4""^@.

#### Explanation:

FULL QUESTION

A particle is projected horizontally from the top of a building in a velocity of ${\text{15 ms}}^{- 1}$. It hits the ground 45m away from the bottom of the building.

• What is the height of the building?
• What is the angle it creates with the horizontal when it hits the ground?

Like with any projectile motion problem, the basic idea is that your particle's trajectory can be broken down into a vertical and a horizontal component.

The same can be said for the particle's initial velocity. However, when you launch a particle horizontally, the nagle it makes with the horizontal is of course ${0}^{\circ}$. This means that you have

$\left\{\begin{matrix}{v}_{0 x} = {v}_{0} \cdot \cos \left({0}^{\circ}\right) = {v}_{0} \\ {v}_{0 y} = {v}_{0} \cdot \sin \left({0}^{\circ}\right) = 0\end{matrix}\right.$

The vertical component of the initial speed will be zero.

So, horizontally, the movement of the particle is not affected by any forces, which means that you can write

$d = {v}_{0 x} \cdot t$, where

$d$ - the distance the particle travels;
$t$ - the time of flight;

Plug your values into this equation to get the time of flight

t = d/v_(0x) = d/v_0 = (45 cancel("m"))/(15cancel("m")/"s") = "3 s"

The particle is in flight for only 3 seconds.

Vertically, the gravitational acceleration will affect the movement of the particle. This implies that vertical component of its velocity will increase as the particle descends from the building.

Taking into account that the vertical displacement is negative, you can write

$- h = {\underbrace{{v}_{0 y}}}_{\textcolor{b l u e}{\text{=0}}} \cdot t - \frac{1}{2} \cdot g \cdot {t}^{2}$

h = 1/2 * g * t^2 = 1/2 * 10"m"/cancel("s"^2) * 3""^2 cancel("s"^2) = color(green)("45 m")#

To get the angle of contact with the ground, you need to determine the vertical component of its final velocity.

Use the equation

${v}_{f y}^{2} = {\underbrace{{v}_{0 y}^{2}}}_{\textcolor{b l u e}{\text{=0}}} + 2 \cdot g \cdot h$

${v}_{f y} = \sqrt{2 \cdot g \cdot h} = \sqrt{2 \cdot 10 \cdot 45} = \text{30 m/s}$

The horizontal component of the velocity is constant,

To get $\theta$, use the definition of the tangent

$\tan \left(\theta\right) = {v}_{0 y} / {v}_{0 x}$

$\tan \left(\theta\right) = \left(30 \cancel{\text{m/s"))/(15cancel("m/s}}\right) = 2$

Therefore, $\theta$ is equal to

$\theta = {\tan}^{- 1} \left(2\right) = \textcolor{g r e e n}{{63.4}^{\circ}}$