# How can the quadratic formula be used to determine where a cannonball shot at a known angle and velocity will land?

Dec 10, 2015

$x = \frac{{v}^{2} \sin \theta \cos \theta}{4 , 9}$

#### Explanation:

From physics principles it can be shown that the equation of motion for the vertical displacement $y$ of any projectile in the y direction as a function of time$t$ may be given in terms if its intial velocity $v$ and projection angle $\theta$, as well as gravitational acceleration $g = 9 , 8 m / {s}^{2}$, as follows

$y = v t \sin \theta - \frac{1}{2} g {t}^{2}$

$= - 4 , 9 {t}^{2} + \left(v \sin \theta\right) t$

This is a quadratic expression in t and clearly the cannonball will land when its vertical height is zero, that is, when

$= - 4 , 9 {t}^{2} + \left(v \sin \theta\right) t = 0$

We may now use the quadratic formula to solve for t in this quadratic equation to obtain
$t = 0 \mathmr{and} t = \frac{v \sin \theta}{4 , 9}$

$t = 0$ is at start, and the latter option is when it lands.

Now in the horizontal (x) direction, the velocity is constant as there is no acceleration in that direction, hence its displacement in the x-direction is given from the definition of velocity as :

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

Hence it will land a distance $x = \left(v \cos \theta\right) \cdot \left(\frac{v \sin \theta}{4 , 9}\right)$ metres away from the projection point.