# Ballistic projectiles

A ballistic projectile is an object moving through the air under the force of gravity only. We give a projectile an initial speed by launching it; after that, it experiences acceleration due to gravity. The horizontal component of its velocity remains constant, but the vertical component decreases at a rate of 9.80 m/s^{2} (on Earth).

In our projectile problems, we assume that

- there is no air resistance;
- gravity is constant;
- the projectile travels in a plane.

All projectile problems can be solved by splitting up original velocity ($\displaystyle \vec{v}_{0}$) into its horizontal and vertical components with

$\displaystyle \vec{v}_{\text{h}} = | \vec{v}_{0} \left \lvert \cos{\theta} \qquad \allowbreak\quad\allowbreak\text{and}\allowbreak\quad\allowbreak \qquad \vec{v}_{\text{1,v}} = \mid \vec{v}_{0} \right \rvert \sin{\theta}$,

where $\displaystyle \theta$ is the direction of the original velocity expressed as an angle above horizontal. These can be used with the equations

$\displaystyle \Delta{} \vec{d}_{\text{h}} = \vec{v}_{\text{h}} \Delta{} t \qquad \allowbreak\quad\allowbreak\text{and}\allowbreak\quad\allowbreak \qquad \Delta{} \vec{d}_{\text{v}} = \vec{v}_{\text{1,v}} \Delta{} t + \frac{1}{2} \vec{a} \left ( \Delta{} t \right )^{2}$.

The duration of the projectile’s flight ($\displaystyle \Delta{} t$) is associated with the vertical part. The *range* (distance to the landing point) is associated with the horizontal part.

Since we physics students don’t like doing any more work than absolutely necessary, we often use more specific, time-saving equations. The first equation tells us how long it takes ($\displaystyle \Delta{} t$) for the projectile to return to its original altitude ($\displaystyle \Delta{} \vec{d}_{\text{v}} = 0$). If we let $\displaystyle g{=} \left \lvert \vec{a}_{\text{g}} \right \rvert$, then

$\displaystyle \Delta{} t = \frac{2 \left \lvert \vec{v}_{0} \right \rvert \sin{\theta}}{g}$,

and the range of this projectile is given by

$\displaystyle \Delta{} \vec{d}_{\text{h}} = \frac{\left \lvert \vec{v}_{0} \right \rvert^{2} \sin{2} \theta}{g}$.

The next equation doesn’t require the projectile to land at its original altitude. It treats the projectile’s altitude as a function of its horizontal position. This allows us to find the projectile’s vertical position at any particular horizontal position and, using the quadratic formula, go the other way as well. Assuming the projectile starts at the origin and passes through a point ($\displaystyle x$, $\displaystyle y$), then

$\displaystyle y = \left ( \tan{\theta} \right ) x - \frac{g}{2 \left \lvert \vec{v}_{0} \right \rvert^{2} \cos^{2}{\theta}} x^{2}$.

The derivation of these three equations is left as an exercise to the reader.

## Example

A soccer ball is kicked at 14.5 m/s [29º a.h.]. Will it make it over a 1.5 m fence 5 m ahead?

When the ball has gone 5 m forward horizontally, $\displaystyle x = 5$ and

$\displaystyle y = \left ( \tan{29} º \right ) \left ( 5 \, \text{m} \right ) - \frac{9.80 \, \text{m/s}^{2}}{2 \left ( 14.5 \, \text{m/s} \right )^{2} \cos^{2}{\left ( 29 º \right )}} \left ( 5 \, \text{m} \right )^{2} = 2.01 \, \text{m}$.

Since
$\displaystyle 2.01 \, \text{m} > 1.5 \, \text{m}$,
the answer is yes—it *will* clear the fence.