# A projectile is shot from the ground at a velocity of 13 m/s and at an angle of (2pi)/3. How long will it take for the projectile to land?

Mar 18, 2018

2.3 seconds

#### Explanation:

Let's breakdown the velocity into x and y parts using trigonometry.

The x axis has a cosine term and the y axis has a sine term, so
${v}_{x} = \cos \left(2 \frac{\pi}{3}\right) \cdot 13 \frac{m}{s} = - \frac{1}{2} \cdot 13 \frac{m}{s} = - 6.5 \frac{m}{s}$
${v}_{y} = \sin \left(2 \frac{\pi}{3}\right) \cdot 13 \frac{m}{s} = \frac{\sqrt{3}}{2} \cdot 13 \frac{m}{s} \approx 11.26 \frac{m}{s}$

We don't care about the x coordinate, because we only want to know how long it takes for the y coordinate to go back to 0.

We can use basic kinematics:
$y = \frac{1}{2} a {t}^{2} + {v}_{0} t \implies y = - \frac{1}{2} g {t}^{2} + {v}_{y} t$

We want to know when this hits the ground, i.e. when $y = 0$:
$y = 0 = - \frac{1}{2} {>}^{2} + {v}_{y} t = - \frac{1}{2} t \cdot \left(g t - 2 {v}_{y}\right)$
which is 0 when $t = 0$ or $\frac{2 {v}_{y}}{g}$

The first makes sense because when we first shoot it off, it is at the ground,

The second solution is
$t = \frac{2 {v}_{y}}{g} \approx \frac{2 \cdot 11.26 \frac{m}{s}}{9.8 \frac{m}{s} ^ 2} \approx 2.3 s$