Correct formula to find the range of a projectile (given angle (possibly negative), initial velocity, initial elevation, and g)

Given:

• $\theta$ (a negative angle)
• $v_0$ (initial velocity)
• $y_0$ (initial height)
• $g$ (acceleration of gravity)

I want to find the range of a projectile (ignoring wind resistance)

Hours of searching have given no useful results. Those that I thought were useful turned out to be incapable of accounting for negative angles. This formula in particular returns a range of 0 if the angle is 0, which simply doesn't make sense.

For example:

• $\theta=-12.5$°
• $v_0=100$m/s
• $y_0=1.65$m
• $g=9.80665$
• What is your question? – Chris2018 Jul 2 '18 at 21:16
• is $\theta$ an angle to the horizontal? – Chris2018 Jul 2 '18 at 21:17

Assuming a projectile is launched at angle $\theta$ from horizontal from a height $y_0$ above the ground, where the ground is assumed to be represented by the line $y = 0.$ The measurement of the angle is such that $\theta = 0$ if the projectile initially is moving horizontally to the right, and otherwise the angle is measured in a counterclockwise direction from "horizontal to the right," so small positive angles are upward and small negative angles are downward.

The objective is to calculate the horizontal distance $d$ from the launch point to the point where the projectile hits the ground, that is, when the $y$ coordinate of the projectile reaches zero.

One way to compute $d$ is first to compute $t_f - t_i,$ the time that elapses from when the projectile is launched until the projectile hits the ground. We can arbitrarily set $t_i = 0.$ If the initial angle $\theta$ were positive, the projectile would first have to rise to the height where it stops rising, and then fall from that height to the ground. The upward velocity at time $t$ after the launch is $$v_y(t) = v_0 \sin\theta - gt.$$ If $t_0$ is the time such that $v_y(t_0) = 0,$ then $$t_0 = \frac{v_0 \sin\theta}{g}.$$ The height of the projectile at that time is $$y(t_0) = y_0 + \frac{v_0^2\sin^2\theta}{2g}.$$ The projectile then falls this far in the interval of time $t_f - t_0,$ that is, $$\frac12 g(t_f - t_0)^2 = y_0 + \frac{v_0^2\sin^2\theta}{2g},$$ and therefore \begin{align} t_f &= t_0 + \sqrt{\frac2g\left(y_0 + \frac{v_0^2\sin^2\theta}{2g}\right)} \\ &= \frac{v_0 \sin\theta}{g} + \sqrt{\frac{2y_0}{g} + \frac{v_0^2\sin^2\theta}{g^2}}. \end{align}

Arbitrarily setting the $x$ coordinate of the starting point to $0,$ the horizontal position of the projectile at any time $t$ is $$x(t) = v_0 t \cos\theta.$$ The desired distance $d$ is simply the $x$ coordinate at time $t_f$ (when the projectile hits the ground), so \begin{align} d &= v_0 t_f \cos\theta \\ &= v_0 \left( \frac{v_0 \sin\theta}{g} + \sqrt{\frac{2y_0}{g} + \frac{v_0^2\sin^2\theta}{g^2}} \right) \cos\theta \\ &= \frac{v_0 \cos\theta}{g} \left( v_0 \sin\theta + \sqrt{2gy_0 + v_0^2\sin^2\theta} \right). \tag1 \end{align}

In the case $y_0 = 0,$ the square root in $(1)$ is just $v_0 \sin\theta,$ and the formula can be simplified to the form $\frac{v_0^2 \sin(2\theta)}{g}$ which is often given as the range of a projectile.

In the case where $\sin\theta \neq 0,$ we can pull a factor of $v_0\sin\theta$ out of the expression in parentheses in $(1)$ to get $$d = \frac{v_0^2\sin\theta\cos\theta}{g} \left( 1 + \sqrt{\frac{2y_0g}{v_0^2\sin^2\theta} + 1} \right).$$

Then, using the fact that $\sin\theta\cos\theta = \frac{\sin(2\theta)}{2},$ we get the formula you found on Wikipedia. But if $\theta = 0$ then $\sin\theta = 0$ and the Wikipedia formula has a division by zero, hence it does not evaluate to zero--it does not evaluate at all.

• Right, accidentally said clockwise instead of counterclockwise. – Jordan T. Knobel-Piehl Jul 2 '18 at 23:24
• Just to confirm I'm doing this right: if theta=0, then d=v0(2y0/g) right? – Jordan T. Knobel-Piehl Jul 2 '18 at 23:44
• d=v0(sqrt(2y0/g)) forgot the radical – Jordan T. Knobel-Piehl Jul 2 '18 at 23:58
• Yes, in that case the distance would simplify to $d=v_0\sqrt{2y_0/g}.$ – David K Jul 3 '18 at 3:50

With horizental velocity $v_x=v_0\cos\theta$ and vertical velocity $v_y=-v_0\sin\theta-gt$ (here $\theta>0$) and integration we see $$x=\int_0^tv_0\cos\theta\,dt=v_0\cos\theta\,t\,\,\,,\,\,\,y=\int_0^t v_y\,dy=-v_0\sin\theta\,t-\frac12gt^2$$ Eliminating $t$ between them gives $$y=-\tan\theta\,x-\frac{g}{2v_0^2\cos^2\theta}x^2$$ the range $R$ is the point $(R,y_0<0)$ and you may continue from here!

Edit: The solution for $h=-y_0$ and $\theta<0$ after solving the equation $$\frac{g}{2v_0^2\cos^2\theta}d^2+\tan\theta\,d-h=0$$ is $$d=-\dfrac{v_0^2}{2g}\sin2\theta\left(1+\sqrt{1+\dfrac{2gh}{v_0^2\sin^2\theta}}\right)$$ which works for $\theta<0$ as you wanted.

You can try the following:

The initial horizontal velocity is $v\cos\theta$, the initial vertical velocity is $-v\sin\theta$ (note that it's positive). Before landing on the field, the coordinates of the rocket follows the following equations:

$$x=v\cos\theta\times t$$ $$y=h-v\sin\theta\times t-\frac{gt^2}{2}$$

The rocket lands at the moment $T$ such that $y(T)=0$. Solving the equation $y(X)=0$, one might be able to choose the root that "good" root for $T$. Put this into $x$ you get the range $x(T)$.

$V_y = 100 \sin(-12.5) = -21.6440$ m/s

$V_x = 100 \cos(-12.5) = 97.6296$ m/s

Solving for the time of flight t

$S = V_y\cdot t + \frac{1}{2}\cdot g\cdot t^2$

$-1.65 = -21.6440t + \frac{1}{2}\cdot -9.80665\cdot t^2$

$4.90333t^2 + 21.6440t - 1.65 = 0$

$t = \frac{-21.6440+\sqrt{21.6440^2-(4\cdot 4.90333\cdot -1.65})}{2\cdot 4.90333}$

$t = .07496$ seconds

Range is $x = V_x\cdot t$

$x = 97.6296\cdot .07496 = 7.3184$ m

In summary: $$x = V_x\cdot (\frac{V_y + \sqrt{V_y^2 - (4\cdot \frac{g}{2}\cdot S)}}{g})$$

Where:

$V_x = 100\cos(-12.5)$

$V_y = 100\sin(-12.5)$

$g = -9.80665$

$S = -1.65$

• my answer was 438m. – Nosrati Jul 2 '18 at 23:09
• The initial height was 1.65 m pointed downwards at an angle of 12.5 degrees? Is that what -12.5 degrees is? – Phil H Jul 2 '18 at 23:19
• The example I gave was -12.5, not 12.5, as in 12.5 degrees downward. I'm looking for a formula that works with negative angles. – Jordan T. Knobel-Piehl Jul 2 '18 at 23:19
• For my method, it creates an inconsistency with negative velocities $(V_x)$ Making $V_0$ negative just shifts the inconsistency somewhere else. – Phil H Jul 2 '18 at 23:42
• To make it consistent, $g, S$ and $V_y$ are all negative (downwards), See edited solution. – Phil H Jul 3 '18 at 0:01