1. State the problem\nYou want to find and draw the trajectory of a ball thrown into the air (projectile motion).\n\n2. Write the model (ignore air resistance)\nLet the launch speed be $v_0$, the launch angle be $\theta$, and take $g$ as the acceleration due to gravity.\n\nThe standard components are:\n$$v_{0x}=v_0\cos\theta$$\n$$v_{0y}=v_0\sin\theta$$\n\n3. Use the parametric equations\nHorizontal motion has constant velocity:\n$$x(t)=v_0\cos\theta\,t$$\nVertical motion has constant acceleration:\n$$y(t)=v_0\sin\theta\,t-\frac{1}{2}gt^2$$\n\n4. Eliminate time to get the trajectory $y$ as a function of $x$\nFrom $x(t)=v_0\cos\theta\,t$, solve for $t$:\n$$t=\frac{x}{v_0\cos\theta}$$\nNow substitute into $y(t)$:\n$$y= v_0\sin\theta\left(\frac{x}{v_0\cos\theta}\right)-\frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2$$\n\nSimplify the first term:\n$$y=\frac{v_0\sin\theta}{v_0\cos\theta}x-\frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2$$\n$$y=\tan\theta\,x-\frac{g}{2v_0^2\cos^2\theta}x^2$$\n\n5. Important rules (learner-friendly)\n- The trajectory is a parabola because $y$ has an $x^2$ term.\n- If $v_0\cos\theta=0$ (meaning $\theta=90^\circ$), then horizontal position doesn’t change and the path is a straight vertical line.\n- The maximum height happens at the top of the parabola; the ball lands when $y=0$ (after launch).\n\n6. Find key points (optional but helpful)\nTime to reach the highest point (when vertical velocity becomes $0$):\n$$t_{\text{top}}=\frac{v_0\sin\theta}{g}$$\nMaximum height:\n$$y_{\max}=v_0\sin\theta\,t_{\text{top}}-\frac{1}{2}g\,t_{\text{top}}^2=\frac{v_0^2\sin^2\theta}{2g}$$\nRange (if it lands back at the launch height $y=0$):\n$$t_{\text{flight}}=\frac{2v_0\sin\theta}{g}$$\n$$R=v_0\cos\theta\,t_{\text{flight}}=\frac{v_0^2\sin(2\theta)}{g}$$\n\n7. What to draw (how to sketch it)\n- Plot a parabola using the equation $y=\tan\theta\,x-\frac{g}{2v_0^2\cos^2\theta}x^2$.\n- Mark the launch point at $(0,0)$.\n- Mark the peak using the $x$ and $y$ at $t_{\text{top}}$: $x_{\text{top}}=v_0\cos\theta\,t_{\text{top}}$ and $y_{\max}$.\n- Mark the landing point at $(R,0)$.\n\n8. Final answer (trajectory equation)\nUnder gravity only (no air resistance), the ball’s trajectory is\n$$y(x)=x\tan\theta-\frac{g}{2v_0^2\cos^2\theta}x^2$$