Here’s a fairly basic kinematics problem that I’ve rarely seen done: consider a projectile with negligible air resistance launched with initial speed *V*_{0} 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;

.

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Tags: Friday Physics, Kinematics, physics, Projectile

This entry was posted on June 13, 2008 at 3:19 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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October 30, 2009 at 1:15 am |

[…] 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 . […]