User talk:Doomed Rasher/Sandbox2

$$\sec^2 \theta = 1 + \tan^2 \theta$$

$$0 = \tan \theta \Delta x - \Delta y + 4.9 \left(\frac{\Delta x}{v\cos \theta}\right)^2 $$

$$0 = \tan \theta \Delta x - \Delta y + 4.9 (1 + \tan^2 \theta) \left(\frac{\Delta x}{v}\right)^2 $$

$$\tan \theta = \frac{\Delta x \pm \sqrt{\Delta x^2 - 96.04\left(\frac{\Delta x}{v}\right)^2(\left(\frac{\Delta x}{v}\right)^2 - \Delta y)}}{9.8\left(\frac{\Delta x}{v}\right)^2}$$

$$\tan \theta = \frac{\Delta x \pm \Delta x\sqrt{1 - 96.04\left(\frac{1}{v}\right)^2(\left(\frac{\Delta x}{v}\right)^2 - \Delta y)}}{9.8\left(\frac{\Delta x}{v}\right)^2}$$

$$\tan \theta = v^2 \frac{1 \pm \sqrt{1 - 96.04\left(\frac{1}{v}\right)^2(\left(\frac{\Delta x}{v}\right)^2 - \Delta y)}}{9.8\Delta x}$$

$$\theta = \arctan \left(v^2 \frac{1 \pm \sqrt{1 - 96.04\left(\frac{1}{v}\right)^2(\left(\frac{\Delta x}{v}\right)^2 - \Delta y)}}{9.8\Delta x}\right)$$