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).

Projectile Trajectory

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 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;


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One Response to “Physics Friday 24: Projectile Range”

  1. Physics Friday 95 « Twisted One 151’s Weblog Says:

    […] not the distance along the ground to the impact point, and so differs from the h=0 case of Physics Friday 24. The only acceleration is due to gravity; so our basic kinematic equations tell us that and . […]

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