# If a projectile is shot at a velocity of 52 m/s and an angle of pi/4, how far will the projectile travel before landing?

Dec 14, 2016

$276 m$

#### 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 takes to its return to the ground after launch be $T$, We would expect two solutions as obviously T=0 is one solution. The total displacement will be zero.

$\left\{\begin{matrix}s = & 0 & m \\ u = & 52 \sin \left(\frac{\pi}{4}\right) = 26 \sqrt{2} & m {s}^{-} 1 \\ v = & \text{not required} & m {s}^{-} 1 \\ a = & - g & m {s}^{-} 2 \\ t = & T & s\end{matrix}\right.$

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

$\therefore 0 = 26 \sqrt{2} T + \frac{1}{2} \left(- g\right) {T}^{2}$
$\therefore \frac{1}{2} g {T}^{2} - 26 \sqrt{2} T = 0$
$\therefore T \left(\frac{g T}{2} - 26 \sqrt{2}\right) = 0$
$\therefore T = 0 , \frac{52 \sqrt{2}}{g}$

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, $t = T$

So we can calculate the horizontal displacement $s$ using $s = u t$

$s = 52 \cos \left(\frac{\pi}{4}\right) T$
$\therefore s = \left(26 \sqrt{2}\right) \left(\frac{52 \sqrt{2}}{g}\right)$
$\therefore s = \frac{2704}{g}$

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

$s = \frac{2704}{9.8} = 275.918 \ldots = 276 m$