# Projectile Motion

## A projectile shot from the origin is aimed at a target located 6000 feet away from the origin horizontally and 3500 feet away vertically.

## What initial velocity is required for the projectile to hit the target?

## From physics and trigonometry we can derive the following parametric equations for the projectile. We will assume the projectile is fired at an angle of 45° from the horizontal. (This angle maximizes the range.) Let v_{0} = initial velocity.

## To hit a target located at the point (6000, 3500), we need to set x = 6000 and y = 3500 and solve the resulting system of equations.

## Solve the x equation for v_{0}:

## Substituting into the y equation:

## Simplify:

## By factoring, or by using the quadratic formula, we get
t = 12.5 or t = -12.5
Eliminate the negative t-value and we find that it took about 12.5 seconds for the projectile to hit the target. But remember, we want to find the initial velocity v_{0}.

## Using the x equation

## where x = 6000 and t = 12.5, we can solve for the initial velocity.

## Solution: To hit the target, the projectile must have an initial velocity of v_{0} = 678.925 ft/sec.