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

Jan 27, 2016

Find the time it takes to reach the maximum height, and use the x component of the displacement equation to find the distance travelled in that time.

#### Explanation:

First we break up the velocity into it's vector components at time 0, ${v}_{x} \left(0\right)$ and ${v}_{y} \left(0\right)$ by using the identities:

${v}_{x} \left(0\right) = v \left(0\right) \cos \left(\theta\right)$
${v}_{y} \left(0\right) = v \left(0\right) \sin \left(\theta\right)$

Since we're presumably ignoring air resistance, the acceleration components are

${a}_{x} = 0$
${a}_{y} = - g$

So our functions for ${v}_{x} \left(t\right)$ and ${v}_{y} \left(t\right)$ becomes

${v}_{x} \left(t\right) = v \left(0\right) \cos \left(\theta\right)$
${v}_{y} \left(t\right) = v \left(0\right) \sin \left(\theta\right) - g \cdot t$

We can find the time it takes to reach the maximum height by setting ${v}_{y} \left(t\right) = 0$ which gives:

$t = v \left(0\right) \sin \frac{\theta}{g}$

Then we use the displacement formula

$x \left(t\right) = x \left(0\right) + {v}_{x} t + \frac{1}{2} {a}_{x} {t}^{2}$

Since ${a}_{x} = 0$, and $x \left(0\right) = 0$ this simplifies to

$x \left(t\right) = {v}_{x} t$

Substitution of ${v}_{x}$ yields

$x \left(t\right) = v \left(0\right) \cos \left(\theta\right) t$

and substituting for t yields:

$x \left(t\right) = \frac{v \left(0\right) \cos \left(\theta\right) v \left(0\right) \sin \left(\theta\right)}{g}$ which simplifies to

$x \left(t\right) = \frac{v {\left(0\right)}^{2} \cos \left(\theta\right) \sin \left(\theta\right)}{g}$ and further simplifies to

$x \left(t\right) = \frac{v {\left(0\right)}^{2} \sin \left(2 \theta\right)}{2 g}$

This gives the range of the projectile at the maximum height.

It's interesting to note that this is exactly half the range equation for a projectile returning to the same height.

$R = \frac{v {\left(0\right)}^{2} \sin \left(2 \theta\right)}{g}$

which is derived using

$y \left(t\right) = y \left(0\right) + {v}_{y} t - \frac{1}{2} g {t}^{2}$

which is the displacement equation for y.