# A projectile is shot at an angle of pi/12  and a velocity of  9 m/s. How far away will the projectile land?

May 24, 2016

Distance to a landing spot is
${V}_{h} \cdot 2 \cdot {V}_{v} / g \approx 4.1326 \left(m\right)$
where
${V}_{h} = 9 \frac{m}{\sec} \cdot \cos \left(\frac{\pi}{12}\right)$
${V}_{v} = 9 \frac{m}{\sec} \cdot \sin \left(\frac{\pi}{12}\right)$

#### Explanation:

Vector of velocity of a projectile should be represented as a sum of two vectors - horizontal and vertical components.
These components are:
${V}_{h} = 9 \frac{m}{\sec} \cdot \cos \left(\frac{\pi}{12}\right) \approx 8.69333 \frac{m}{\sec}$
${V}_{v} = 9 \frac{m}{\sec} \cdot \sin \left(\frac{\pi}{12}\right) \approx 2.32937 \frac{m}{\sec}$

While the vertical component brings the projectile up, after which it falls down, the horizontal component pulls it forward during the same time.

So, to find the distance to a landing spot, we need the time the projectile is flying.
On its way up the projectile's vertical component of the velocity is diminishing every second by $g \frac{m}{{\sec}^{2}} \approx 9.8 \frac{m}{\sec} ^ 2$.
Therefore, with initial vertical component of ${V}_{v} = 9 \frac{m}{\sec} \cdot \sin \left(\frac{\pi}{12}\right)$ and a loss of $9.8 \frac{m}{\sec} ^ 2$ every second, the projectile will continue rising for
${V}_{v} / g \approx \frac{9 \sin \left(\frac{\pi}{12}\right)}{9.8} \approx 0.23769 \left(\sec\right)$

The same time is needed by the projectile to come back to earth, which brings the total time to go up and down to
$t = 0.23769 \cdot 2 = 0.47538 \left(\sec\right)$

During the same vertical component was dragging the projectile forward with a speed ${V}_{h}$.

So, the distance covered is
${V}_{h} \cdot 0.47538 = 9 \cos \left(\frac{\pi}{12}\right) \cdot 0.47538 \approx 4.1326 m$