1. **State the problem.** A ball is thrown into the air (no air resistance). We want the **trajectory** $y(x)$ (the path) and we want a picture to help you understand. 2. **Write the projectile formulas.** Let the ball be launched from $(0,0)$ with speed $v_0$ at angle $\theta$ above the horizontal. Horizontal velocity stays constant: $$v_x=v_0\cos\theta$$ Vertical velocity changes due to gravity $g$: $$v_y(t)=v_0\sin\theta-gt$$ 3. **Write position as a function of time.** Horizontal position: $$x(t)=v_0\cos\theta\,t$$ Vertical position: $$y(t)=v_0\sin\theta\,t-\frac{1}{2}gt^2$$ 4. **Eliminate time to get $y$ as a function of $x$.** From $x(t)=v_0\cos\theta\,t$, solve for $t$: $$t=\frac{x}{v_0\cos\theta}$$ Substitute into $y(t)$: $$y=v_0\sin\theta\left(\frac{x}{v_0\cos\theta}\right)-\frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2$$ 5. **Simplify carefully (showing cancelations).** First term simplification: $$y=\frac{v_0\sin\theta}{v_0\cos\theta}x-\frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2$$ Show cancelation: $$y=\frac{\cancel{v_0}\sin\theta}{\cancel{v_0}\cos\theta}x-\frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2$$ So the first term becomes $x\tan\theta$: $$y=x\tan\theta-\frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2$$ Now simplify the squared part: $$y=x\tan\theta-\frac{1}{2}g\frac{x^2}{v_0^2\cos^2\theta}$$ 6. **Final trajectory equation.** $$\boxed{y(x)=x\tan\theta-\frac{g}{2v_0^2\cos^2\theta}x^2}$$ This is a **parabola** opening downward. 7. **Why this makes sense (important rules).** - The horizontal motion is linear in time because horizontal acceleration is $0$. - The vertical motion is a quadratic in time because gravity makes vertical acceleration $-g$. - When you eliminate time, the result becomes a quadratic in $x$, so the path is a parabola. 8. **(Optional) Key vertex/maximum height idea.** For $y(x)=ax^2+bx$, the vertex occurs at $x_{\max}= -\frac{b}{2a}$. Here $a=-\frac{g}{2v_0^2\cos^2\theta}$ and $b=\tan\theta$, so: $$x_{\max}= -\frac{\tan\theta}{2\left(-\frac{g}{2v_0^2\cos^2\theta}\right)}=\frac{v_0^2\cos^2\theta\,\tan\theta}{g}$$ (That simplifies further using $\tan\theta=\frac{\sin\theta}{\cos\theta}$ if you want.) 9. **Use the formula.** To draw the trajectory for your specific numbers, plug in your $v_0$, $\theta$, and $g$ into $y(x)$ above, then compute a few points $(x,y)$ and sketch the parabola.