# A projectile is shot from the ground at a velocity of 29 m/s and at an angle of (2pi)/3. How long will it take for the projectile to land?

Jun 10, 2016

$x = 74.24 \text{ m}$

#### Explanation: $\text{The velocity of projectile can be split two components as vertical and horizontal}$

$\text{the vertical component can be calculated using the formula:}$
${v}_{y} = {v}_{i} \cdot \sin \alpha - g \cdot t \text{ "figure" } 1$ $\text{you are seeing how the vertical component of velocity is changing }$

$\text{please note that the vertical component of velocity " v_y" is zero at maximum height}$

$0 = {v}_{i} \cdot \sin \alpha - g \cdot t$
$g \cdot t = {v}_{i} \cdot \sin \alpha$
$t = \frac{{v}_{i} \cdot \sin \alpha}{g} \text{ time elapsed to the maximum height}$

$\text{2*t gives us the traveling time}$

$\textcolor{red}{{t}_{t} = \frac{2 \cdot {v}_{i} \cdot \sin \alpha}{g}}$ $\text{the object is conveyed by the horizontal component of velocity "v_x" figure 3}$

$\text{the horizontal component of velocity is calculated by:}$

$\textcolor{g r e e n}{{v}_{x} = {v}_{i} \cdot \cos \alpha}$
$\text{the horizontal component of velocity doesn't change}$

$\text{and doesn't depend on time}$ $\text{the horizontal position of object can be calculated by:}$

$x = \textcolor{g r e e n}{{v}_{x}} \cdot \textcolor{red}{t}$

$x = \textcolor{g r e e n}{{v}_{i} \cdot \cos \alpha} \cdot \textcolor{red}{t}$

$\text{now; we can write } \textcolor{red}{t = {t}_{t}}$

$x = \frac{{v}_{i} \cdot \cos \alpha \cdot \left(2 \cdot {v}_{i} \cdot \sin \alpha\right)}{g}$

$\text{so } 2 \cdot \sin \alpha \cdot \cos \alpha = \sin \left(2 \alpha\right)$

$x = \frac{{v}_{i}^{2} \cdot \sin \left(2 \alpha\right)}{g}$

$\text{where "v_i=29" m/s"" } \alpha = \left(\frac{2 \pi}{3}\right)$

$x = \frac{{29}^{2} \cdot \sin \left(2 \cdot \frac{2 \pi}{3}\right)}{9.81}$

$x = \frac{841 \cdot 0.866}{9.81}$

$x = 74.24 \text{ m}$