# A projectile is shot at a velocity of 6 m/s and an angle of pi/12 . What is the projectile's maximum height?

Feb 23, 2016

$S = {\left(V \cdot \sin \left(\phi\right)\right)}^{2} / \left(2 g\right)$
where
$V = 6 \frac{m}{\sec}$ - initial speed,
$\phi = \frac{\pi}{12}$ - angle to horizon of launching and
$g = 9.8 \frac{m}{\sec} ^ 2$ - gravitational constant

The answer (approximately) is $0.123 m$

#### Explanation:

Initial velocity of $V = 6 \frac{m}{\sec}$ directed at an angle $\phi = \frac{\pi}{12}$ to horizon can be represented as a sum two vectors:

${V}_{v} = V \cdot \sin \left(\phi\right)$ directed vertically up and
${V}_{h} = V \cdot \cos \left(\phi\right)$ directed horizontally.

It's the vertical component ${V}_{v}$ that drives the projectile up against the constant gravity deceleration $g = 9.8 \frac{m}{\sec} ^ 2$

Since every second the projectile looses its vertical component by $g$, we can calculate the time until its vertical component becomes zero (the top of trajectory):
$t = {V}_{v} / g$

Knowing time to the top and initial vertical velocity, the length is calculated by a formula
$S = {V}_{v} \cdot t - \frac{g \cdot {t}^{2}}{2} = {V}_{v}^{2} / g - \frac{g \cdot {V}_{v}^{2}}{2 {g}^{2}} = {V}_{v}^{2} / \left(2 g\right)$

Incidentally, this is equal to a time $t = {V}_{v} / g$ multiplied by an average speed, which under constant deceleration from ${V}_{v}$ to $0$ equals to ${V}_{v} / 2$.

Substituting the real values for ${V}_{v}$ and $g$,
$S = \frac{{6}^{2} \cdot {\sin}^{2} \left(\frac{\pi}{12}\right)}{2 \cdot 9.8} = \frac{18 \cdot \left(2 {\sin}^{2} \left(\frac{\pi}{12}\right)\right)}{19.6} =$
$= 0.92 \left(1 - \cos \left(\frac{\pi}{6}\right)\right) = 0.92 \cdot \left(1 - \frac{\sqrt{3}}{2}\right) =$
$0.92 \cdot 0.134 = 0.123 \left(m\right)$