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Maximizing distance traveled by a thrown object

A puzzle from mathpuzzle.com, November 11, 2004.

Ed Pegg Asked

You throw a ball over a flat field as hard as you can. Easy: what upward angle maximizes the distance away that the ball lands? Harder: what upward angle maximizes the distance that the ball travels?

Basic Equations

This is a basic dynamics problem where details like wind resistance and the curvature of the earth are meant to be ignored. Suppose you throw the ball with an initial speed $V_{0}$ at an initial angle $θ$ above the horizontal. This gives it an initial upward component of velocity $V = V_{0} sin (θ)$ and an unchanging forward component of velocity of $H = V_{0} cos (θ)$ . The vertical velocity is reduced by the acceleration due to gravity; that is $\frac{d V}{d t} = - g$ and therefore $V (t) = V_{0} sin (θ) - g t$ .

Due to conservation of energy, the ball reaches the ground at the same speed, but downward, that it had when you threw it. So if the final time is t=F, $V (F) = - V_{0} sin (θ) = V_{0} sin (θ) - g F$ so $F = 2 (\frac{V_{0}}{g}) sin (θ)$ .

Part 1 Solution

Since it travels the same horizontal speed at all times, to compute the distance A away from you where the ball lands, you simply need to multiply the time it is in the air times the horizontal velocity. Thus maximize $A = F H = 2 (\frac{V_{0}^{2}}{g}) sin (θ) cos (θ) = (\frac{V_{0}^{2}}{g}) sin (2 θ)$ . $\frac{d F H}{d t h η} = 2 (\frac{V_{0}^{2}}{g}) cos (2 θ)$ which is zero when cos(2 theta) is zero at $θ = \frac{π}{4}$ .

Part 2 Solution

The distance traveled D is a bit harder to compute. This is the integral of the full speed S (considering both components of the velocity) from 0 to F. $S (t) = \sqrt{V {(t)}^{2} + H^{2}} = \sqrt{V_{0}^{2} {sin}^{2} (θ) + g^{2} t^{2} - 2 V_{0} g t sin (θ) + V_{0}^{2} {cos}^{2} (θ)}$ .

Since ${sin}^{2} (θ) + {cos}^{2} (θ) = 1$ , this reduces to $S (t) = \sqrt{V_{0}^{2} + g^{2} t^{2} - 2 V_{0} g t sin (θ)}$ . Then the distance traveled is $\int_{0}^{F} S (t) d t$ . How do you integrate this?

The Standard Math Tables provide a solution. Let $S (t) = \sqrt{X}$ , $X = a + b t + c t^{2}$ , $a = V_{0}^{2}$ , $b = - 2 V_{0} g sin (θ)$ , and $c = g^{2}$ . Then $q = 4 a c - b^{2} = 4 V_{0}^{2} g^{2} (1 - {sin}^{2} (θ)) = 4 V_{0}^{2} g^{2} {cos}^{2} (θ)$ , and $k = 4 \frac{c}{q} = \frac{1}{V_{0}^{2} {cos}^{2} (θ)}$ , then $\int \sqrt{X} d t = (2 c t + b) \frac{\sqrt{X}}{4 c} + (\frac{1}{2 k}) \int \frac{d t}{\sqrt{X}}$ .

The math tables also provide a solution for the new integral. Actually, it provides three of them, but the one that fits best here, for $c > 0$ as it is in this case, is $\int \frac{d t}{\sqrt{X}} = (\frac{1}{\sqrt{c}}) log (\sqrt{X} + t \sqrt{c} + \frac{b}{2 \sqrt{c}})$ .

Substituting in the variables from the problem, $\int S (t) d t = (g t - V_{0} sin (θ)) S \frac{t}{2 g} + \frac{V_{0}^{2} {cos}^{2} (θ)}{2 g} log (S (t) + g t - V_{0} sin (θ))$ . Then we need to evaluate this at $F = 2 (\frac{V_{0}}{g}) sin (θ)$ and 0. Remember from above that $S (0) = S (F) = V_{0}$ .

$D = \frac{(2 V_{0} sin (θ) - V_{0} sin (θ)) V_{0}}{2 g} + \frac{V_{0}^{2} {cos}^{2} (θ)}{2 g} log (V_{0} + 2 V_{0} sin (θ) - V_{0} sin (θ)) + (V_{0} sin (θ)) \frac{V_{0}}{2 g} - \frac{V_{0}^{2} {cos}^{2} (θ)}{2 g} log (V_{0} - V_{0} sin (θ))$

$D = \frac{V_{0}^{2} sin (θ)}{g +} \frac{V_{0}^{2} {cos}^{2} (θ)}{2 g} (log (V_{0} + V_{0} sin (θ)) - log (V_{0} - V_{0} sin (θ)))$

$D = \frac{V_{0}^{2} sin (θ)}{g +} \frac{V_{0}^{2} {cos}^{2} (θ)}{2 g} log (\frac{1 + sin (θ)}{1 - sin (θ)})$

The arrangement inside the logarithm suggests the application of trig identities. Multiply numerator and denominator by $1 + sin (θ)$ , leading to $log (\frac{{(1 + sin (θ))}^{2}}{{cos}^{2}} (θ)) = 2 log (\frac{1 + sin (θ)}{cos (θ)})$ .

$D = \frac{V_{0}^{2} sin (θ)}{g +} \frac{V_{0}^{2} {cos}^{2} (θ)}{g} log (\frac{1 + sin (θ)}{cos (θ)})$

The angle at which the maximum in the distance occurs is then the solution of $\frac{d D}{d t h η} = 0$

$\frac{d D}{d t h η} = \frac{V_{0}^{2} cos (θ)}{g -} \frac{2 V_{0}^{2} cos (θ) sin (θ)}{g log (\frac{1 + sin (θ)}{cos (θ)})} + \frac{V_{0}^{2} {cos}^{2} (θ)}{g} \frac{cos (θ)}{1 + sin (θ)} ({sec}^{2} (θ) + sec (θ) tan (θ)) = 0$

$cos (θ) - 2 cos (θ) sin (θ) log (\frac{1 + sin (θ)}{cos (θ)}) + {cos}^{2} (θ) \frac{cos (θ)}{1 + sin (θ)} ({sec}^{2} (θ) + sec (θ) tan (θ)) = 0$

$cos (θ) - 2 cos (θ) sin (θ) log (\frac{1 + sin (θ)}{cos (θ)}) + \frac{cos (θ)}{1 + sin (θ)} (1 + sin (θ)) = 0$

$cos (θ) - 2 cos (θ) sin (θ) log (\frac{1 + sin (θ)}{cos (θ)}) + cos (θ) = 0$

$2 cos (θ) = 2 cos (θ) sin (θ) log (\frac{1 + sin (θ)}{cos (θ)})$

$1 = sin (θ) log (\frac{1 + sin (θ)}{cos (θ)})$

This can be solved numerically, to find $θ$ is about 0.985515 radians or 56.4658 degrees. And $D (θ) = 1.19968 \frac{V_{0}^{2}}{g}$ . Compare this with the distance at the 45 degree angle throw, where $D = 1.14779 \frac{V_{0}^{2}}{g}$ .