# If a projectile is shot at a velocity of 5 m/s and an angle of pi/4, how far will the projectile travel before landing?

Dec 6, 2016

$2.55 m$

#### Explanation:

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

$\left.\begin{matrix}v = u + a t & \text{ where " & s="displacement "(m) \\ s=ut+1/2at^2 & \null & u="initial speed "(ms^-1) \\ s=1/2(u+v)t & \null & v="final speed "(ms^-1) \\ v^2=u^2+2as & \null & a="acceleration "(ms^-2) \\ s=vt-1/2at^2 & \null & t="time } \left(s\right)\end{matrix}\right.$

Vertical Motion
Motion under constant acceleration due to gravity, applied vertically upwards

Let the total time that the projectile is in the air be $T$, then then total vertical displacement will be $0$, we would expect two solutions, one will be $t = 0$

$\left\{\begin{matrix}s = & 0 & m \\ u = & \left(5\right) \sin \left(\frac{\pi}{4}\right) = \frac{5}{2} \sqrt{2} & m {s}^{-} 1 \\ v = & \text{not required} & m {s}^{-} 1 \\ a = & - g & m {s}^{-} 2 \\ t = & T & s\end{matrix}\right.$

So we can calculate $T$ using $s = u t + \frac{1}{2} a {t}^{2}$

$\therefore 0 = \frac{5}{2} \sqrt{2} T - \frac{1}{2} g {T}^{2}$
$\therefore \frac{1}{2} T \left(5 \sqrt{2} - g T\right) = 0$
$\therefore T = 0 , \frac{5 \sqrt{2}}{g}$

Horizontal Motion
Under constant speed (NB we can still use "suvat" equation with a=0). Thje projectile will be in in the air for the same time, $t = T$

$s = \left(5\right) \cos \left(\frac{\pi}{4}\right) T$
$\therefore s = \frac{5 \sqrt{2}}{2} T$
$\therefore s = \frac{5 \sqrt{2}}{2} \cdot \frac{5 \sqrt{2}}{g}$
$\therefore s = \frac{25}{g}$

So using $g = 9.8 m {s}^{-} 2$ we have.

$s = \frac{25}{9.8} = 2.55102 \ldots = 2.55 m$