# A projectile is shot from the ground at an angle of pi/4  and a speed of 2/5 ms^-1. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Dec 8, 2016

$8.1 m m$

#### 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 maximum point be $T$, at the point the velocity will momentary be zero.

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

So we can calculate $T$ using $v = u + a t$

$\therefore 0 = \frac{1}{5} \sqrt{2} - g T$
$\therefore g T = \frac{1}{5} \sqrt{2}$
$\therefore T = \frac{\sqrt{2}}{5 g}$

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 $s$ using $s = u t$

$s = \frac{2}{5} \cos \left(\frac{\pi}{4}\right) T$
$\therefore s = \frac{1}{5} \sqrt{2} T$
$\therefore s = \frac{1}{5} \sqrt{2} \frac{\sqrt{2}}{5 g}$
$\therefore s = \frac{2}{25 g}$

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

$s = \frac{2}{245} = 0.008163 \ldots = 0.0081 m$, or $8.1 m m$