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 v0 = 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 v0:
Substituting into the y equation:
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 v0.
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 v0 = 678.925 ft/sec.