# A projectile is shot from the ground at an angle of pi/12  and a speed of 2/3 m/s. When the projectile is at its maximum height, what will its horizontal distance from the starting point be?

Jul 2, 2016

${x}_{h \max} = 0.011 m$ to 3d.p

#### Explanation:

A wee disclaimer: I'm going to neglect air resistance in this answer, but if you want I can include it. I'm also going to go into a bit of depth and derive most things, it might be going a bit overboard but I certainly like to have an idea about where things are coming from when I use them so i felt I'd include it.

We are going to split the motion of the projectile into the horizontal and vertical directions in order to examine this problem. Our coordinate system will treat upwards as positive y-direction and right as positive x-direction.

Starting with the vertical direction seems sensible, since we need to know how long it'll take to get to it's maximum height. Without air resistance, the only force acting on the projectile is gravity, which will apply an acceleration in the negative y-direction. From Newton's second law, the equation of motion for the projectile is

$m \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = - m g$

$\cancel{m} \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = - \cancel{m} g$

Pretty simple separable differential equation, integrating gives:

$\frac{\mathrm{dy}}{\mathrm{dt}} = - g t + {C}_{1}$

$\frac{\mathrm{dy}}{\mathrm{dt}}$ is of course the vertical velocity of the projectile and we shall denote this ${v}_{y} \left(t\right)$ with initial velocity ${v}_{y 0}$ at t = 0s. Using this initial condition we get that

${v}_{y} \left(0\right) = {v}_{y 0} = {C}_{1}$

So this means that:

${v}_{y} \left(t\right) = {v}_{y 0} - g t$

The particles vertical velocity will be zero when it reaches it's maximum height, so setting ${v}_{y} = 0$ will give the time at which this height is reached.

ie ${t}_{h \max} = \frac{{v}_{y 0}}{g}$

So we've got the time, we just need to compute the distance. Again using Newton's second law, the equation of motion in the x direction will be

$m \frac{{d}^{2} x}{{\mathrm{dt}}^{2}} = 0$

as there are no forces (in this idealised example) acting in the x direction. Integrating gives

$\frac{\mathrm{dx}}{\mathrm{dt}} = {C}_{2}$

Because there are no forces, the velocity in the horizontal direction is constant, so ${v}_{x} \left(t\right) = {v}_{x 0}$. Another simple, separable which, upon integration yields:

$x \left(t\right) = {v}_{x 0} t + {C}_{3}$

$x \left(0\right) = 0 \implies {C}_{3} = 0$ and leaves that $x \left(t\right) = {v}_{x 0} \cdot t$

So now we have an expression for the distance and an expression for the time, so let's combine them!

${x}_{h \max} = {v}_{x 0} \cdot \left(\frac{{v}_{y 0}}{g}\right)$

The velocity given in the question is the vector sum of the horizontal and vertical components, using trigonometry we can obtain that
${v}_{y 0} = v \cdot \sin \theta$
${v}_{x 0} = v \cdot \cos \theta$

So ${x}_{h \max} = \frac{\frac{2}{3} \cdot \cos \left(\frac{\pi}{12}\right) \cdot \frac{2}{3} \cdot \sin \left(\frac{\pi}{12}\right)}{9.81}$

Giving ${x}_{h \max} = 0.011 m$ to 3d.p