1. The problem is to find the trajectory of a ball thrown into the air. 2. The trajectory of a projectile under gravity is given by the equations: $$x = v_0 \cos(\theta) t$$ $$y = v_0 \sin(\theta) t - \frac{1}{2} g t^2$$ where $v_0$ is the initial velocity, $\theta$ is the launch angle, $g$ is the acceleration due to gravity (approximately 9.8 m/s²), $t$ is time, $x$ is horizontal position, and $y$ is vertical position. 3. These equations come from decomposing the initial velocity into horizontal and vertical components and applying the equations of motion under constant acceleration. 4. To find the trajectory, eliminate $t$ from the equations: From $x = v_0 \cos(\theta) t$, we get $$t = \frac{x}{v_0 \cos(\theta)}$$ Substitute into $y$: $$y = v_0 \sin(\theta) \frac{x}{v_0 \cos(\theta)} - \frac{1}{2} g \left(\frac{x}{v_0 \cos(\theta)}\right)^2$$ Simplify: $$y = x \tan(\theta) - \frac{g x^2}{2 v_0^2 \cos^2(\theta)}$$ 5. This is the equation of the parabolic trajectory of the ball. 6. The ball follows a curved path shaped like a parabola, rising to a maximum height and then falling back down. 7. The maximum height and range depend on $v_0$ and $\theta$. 8. Below is a simple SVG diagram illustrating the parabolic trajectory of the ball thrown at an angle $\theta$ with initial velocity $v_0$.