Physics Friday 24: Projectile Range

Here’s a fairly basic kinematics problem that I’ve rarely seen done: consider a projectile with negligible air resistance launched with initial speed V0 at an angle θ above the horizontal () from a point a height h above the ground (this is the rare part).

What value of θ gives the projectile its maximum horizontal range R, and what is that range?

Choosing the point on the ground beneath the launching point as the origin, we have the initial x-component of velocity is , and the initial y-component is . Thus, the position of the projectile as a function of time is:
 and .
We can solve the first equation for t and substitute into the second to get an equation for the trajectory:


Now, for later convienience, let us define the value . Note that L has units of length. With L, our trajectory is written
.

To find the horizontal range, we want to find  where . This is finding the positive root of the quadratic , which, using the quadratic formula, is


Now, to find the theta where this is a maximum, we first take the derivative with respect to θ:
.
Next, we find the value of θ for which this is zero. We note that the fraction on the right hand side above is zero when its numerator is zero, and so we have
.
Solving,

Solving for the term with the square root and squaring both sides,


So we see that when h=0, we have , (=45°) as is generally found in most introductory physics texts. Note that as h increases, the angle decreases; the higher you launch your projectile, the more horizontally you will want to angle it to achieve maximum range ( as ).

Now, we just need to plug this θ into our value for R. We have , and thus (using a little trig),
.
Plugging these into
,
we find
,
or in terms of the original parameters;
.