### introduction

A recent homework issue on the forums involved maximizing the horizontal reach of a projectile, with the launch location being a fixed height above the ground that the projectile would eventually impact. There were a number of interesting solution methods, so this article aims to introduce all of them, as each method represents a different approach to the problem and each is instructive in itself.

### The problem

A cannon sits on an airplane at a height ## h ## above the ground and fires a grenade at an initial speed of ## v_0 ##. At what angle ## theta_0 ## does the grenade have to be fired in order to achieve the maximum range ## R _ { text {max}} ## along the ground?

Hope I won’t mind the OP (original poster) of the above problem including its problem diagram: ### Parametric equations

The above motion can be represented by the following pair of parametric equations: \$\$ x = R = v_0 cos ( theta) t \$\$ and \$\$ y = v_0 sin ( theta) t-½gt ^ 2. \$\$ if if we replace t in the second equation and set y = -h, we get: \$\$ – h = R tan ( theta) – frac {gR ^ 2 sec ^ 2 ( theta)} {2 {v_0} ^ 2} \$\$

### Method 1: Implicit Differentiation

From PF user @Perok:

Put ## u = tan theta ## and rewrite the equation as follows:

\$\$ frac {gR ^ 2} {2v ^ 2} (1 + u ^ 2) – Ru = h \$\$ Then differentiate implicitly wrt ## u ## and set ## frac {dR} {du} = 0 ##, type: \$\$ R = frac {v ^ 2} {gu} \$\$ Then plug that into the first equation:
\$\$ h = frac {gR ^ 2} {2v ^ 2} (1 + u ^ 2) – Ru = frac {v ^ 2} {2gu ^ 2} (1 + u ^ 2) – frac {v ^ 2} {g} = frac {v ^ 2} {2g} ( frac {1} {u ^ 2} – 1) \$\$ to give: \$\$ u ^ 2 = frac {1} {1 + 2gh / v ^ 2} text {and} frac 1 u = sqrt {1 + frac {2gh} {v ^ 2}} \$\$ Then you get ## R ## directly from:
\$\$ R = frac {v ^ 2} {gu} = frac {v ^ 2} {g} sqrt {1 + frac {2gh} {v ^ 2}} = frac {v} {g} sqrt {v ^ 2 + 2gh} \$\$
And of course: \$\$ tan theta = frac {v} { sqrt {v ^ 2 + 2gh}} \$\$
Note that you can also insert ## R = frac {v ^ 2} {g tan theta} ## into the first equation and get an equation for ## sin theta ##, since the trigger functions cancel .

### Method 2: “Imaginary Slope” method

In this method, we draw a line from which the projectile is fired up to its point of impact on the flat ground, which is a fixed vertical distance h lower than the point of launch. Then we treat the problem with equations that apply to projectiles that are fired or (in this case) down a slope. The mathematical “hard yards” for this approach are presented to us in this section of a Kinematics Fundamentals website.

We focus on the derived equation for the distance along a slope and find that maximizing this distance (R ‘) also maximizes the horizontal distance (R) traveled, when the vertical distance is fixed:

\$\$ R ‘= frac {{v_0} ^ 2} {g cos ^ 2 alpha} left ( sin (2 theta + alpha) + sin alpha right), \$\$ where ## alpha ## is the angle of our “imaginary” slope with ## sin alpha = frac {h} {R ‘}, ## and ## R’ = h ^ 2 + R ^ 2 ##. We can see that this expression is maximized when \$\$ 2 theta + alpha = frac { pi} {2} implies theta_ {max} = frac { pi} {4} – frac { alpha} {2}. \$\$ Under these circumstances we get \$\$ R ‘_ {max} = frac {{v_0} ^ 2} {g cos ^ 2 alpha} left (1+ sin alpha right) = frac { {v_0} ^ 2} {g (1- sin alpha)}, \$\$ and replacing ## R ‘_ {max} = h / sin alpha ## we solve the equation: \$ \$ frac { h} { sin alpha} = frac {{v_0} ^ 2} {g (1- sin alpha)} \$\$ get: \$\$ sin alpha = frac {hg} {{v_0} ^ 2 + hg}. \$\$ Finally we get the projection angle for the maximum range: \$\$ theta_ {max} = frac { pi} {4} – frac { alpha} {2}. All that remains to be done is to show that this result is the same as the previous one obtained by implicit differentiation. \$\$ tan theta_ {max} = tan left ( frac { pi} {4} – frac { alpha} {2} right) = frac {1- sin alpha} { cos alpha} = sqrt { frac {1- sin alpha} {1+ sin alpha}}. \$\$ Replace ## sin alpha = frac {hg} {hg + {v_0} ^ 2} ##, we get the appropriate result: \$\$ tan theta_ {max} = frac {v_0} { sqrt {{v_0} ^ 2 + 2gh}} \$\$ We should also check our result for the maximum horizontal range: \$\$ R_ {max} = R ‘_ {max} cos alpha = h cot alpha = frac {hv_0 sqrt {{v_0} ^ 2 + 2gh}} {hg} = frac {v_0 sqrt {{v_0} ^ 2 + 2gh}} {g}. \$\$

### Method 3: using vectors

The “hard yards” for this deceptively simple but very elegant method are covered in a PF Insights article entitled “How to Master Projectile Movement Without Quadratics” by PF user @kuruman. I say “deceptively simple” because the approach requires a change in the mindset that one is used to in conventional projectile kinematics. This writer found (at least) that it took some getting used to. But once you understand the theory, the problem of maximizing the horizontal area becomes an almost trivial “consequence”!

We’ll start with a slightly modified version of @kuruman’s vector diagram. The diagram shows the vector relationship ## vec v _ {! F} = vec v_0 + vec g ~ t _ {! F}. ## Winkel ## theta ## is the projection angle and angle ## theta_c ## is complementary to ## theta ## since triangle ADB is a right triangle. At point D, the projectile has zero vertical velocity, which means that it has reached its highest point. The meaning of the right angle marked at vertex A will be explained a little later.

The horizontal projectile area R relates to the vector cross product of the initial and final velocities: \$\$ vec v_0 times vec v _ {! F} = vec v_0 times ( vec v_0 + vec g ~ t _ {! F}) = vec v_0 times vec g ~ t _ {! f} \$\$ \$\$ vec v_0 times vec g ~ t _ {! f} = v_ {0 perp} ~ g ~ t_ {! f} = (v_ {0 x} ~ t _ {! f}) g = Rg. \$\$ Hence \$\$ R = frac {| vec v_0 times vec v _ {! F} |} {g}. \$\$ This vector cross product is clearly maximized if the angle between ## vec v_0 # # and ## vec v_f ## is a right angle. This is the situation shown in the diagram above, which shows a right angle at vertex A. The geometry should show why we marked the vertex C of the speed triangle with the angle ## theta ##. Then:

\$\$ R_ {max} = frac {| vec v_0 || vec v_f |} {g} \$\$ and \$\$ tan theta_ {max} = frac {| vec v_0 |} {| vec v_f |}. \$\$ Sure ## | vec v_0 | = v_0 ## and – for a projectile that has fallen a fixed vertical distance h below its starting point (as is the case with this problem) – ## | v_f | = sqrt {v_0 ^ 2 + 2gh} ##. If we plug these values ​​into the equations above, we again get exactly the same set of solutions obtained by the other two methods above.

We should add a brief explanation that since ## {v_0} ^ 2 = {v_ {0x}} ^ 2+ {v_ {0y}} ^ 2 ##, it follows that ## {v_f} ^ 2 = { v_x} ^ 2 + {v_y} ^ 2 + 2gh = {v_0} ^ 2 + 2gh ##.

### Summary and conclusion

In this article we have examined the problem of (horizontal) range maximization for a projectile that is fired at a fixed vertical distance h above the ground where it will eventually impact. Three different techniques, each with their own set of mathematical operations, have been described, but they still lead to the same set of solutions for the maximum horizontal range R and the corresponding starting angle.

The first method applied uses a carefully chosen trigonometric substitution which greatly simplifies the solution through a calculus-based method. The second solution “draws” an imaginary inclination and then uses the kinematics of projectiles on an inclined plane. Finally, Method 3 is noteworthy in that it demonstrates the “evergreen” character of Euclidean geometry. Similar triangles in conjunction with vector cross products provide a solution that accurately illustrates an essential physical aspect of the maximization problem, namely that range maximization occurs when the initial and final velocities are at right angles to one another.

### thanksgiving

Thanks to PF user @brotherbobby for reporting the range maximization issue in the first place.

This article was effectively co-authored by PF mentors @Perok and @kuruman who contributed solution methods 1 and 3, respectively.

Further thanks to PF users @ vanhees71, @dextercioby, and @jedishrfu who all took the time to review and / or comment on the article when it was originally posted on the Insights development forum. And of course to @Greg who took care of the little detail of the actual release!

The Geogebra online utility was used to create the vector diagram graphic in this article.

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