# If a projectile is shot at a velocity of 2 m/s and an angle of pi/4, how far will the projectile travel before landing?

Jan 5, 2016

it will land about 0.638 m far.

#### Explanation:

We have to write down the equations of motion:

$x = {x}_{0} + v \cos \theta t$
$y = {h}_{0} + v \sin \theta t + \frac{a {t}^{2}}{2}$

where $v$ is the velocity of the projectile, $\theta$ is the angle of the trajectory, ${h}_{0}$ the initial quota of the projectile, ${x}_{0}$ its initial position.

We have:

• ${h}_{0} = 0$ and ${x}_{0} = 0$;
• $v = 2 \setminus m {s}^{- 1}$
• $\theta = \frac{\pi}{4}$
• $a = - g \setminus \approx - 9.81 \setminus m {s}^{- 2} .$

We have $v$ and $t h e t h a$: we have to find t from the first equation and substitute its value in the second one.
When the projectile lands, it reaches the ground so we have y=0 at that moment.
The equations of motion at the landing are:

$x = v \cos \theta t$
$0 = v \sin \theta t - \frac{{>}^{2}}{2}$

from the first equation $t = \frac{x}{v \cos \theta}$.
The other becomes:
$v \sin \theta \cdot \frac{x}{v \cos \theta} - \frac{g}{2} \cdot {\left(\frac{x}{v \cos \theta}\right)}^{2} =$

Performing the calculation we eventually have:
$x = \sqrt{\frac{2 {v}^{2} \cos \theta \sin \theta}{g}} \approx 0.638 \setminus m$