# Nakafa Learning Content > For AI agents: use [llms.txt](https://nakafa.com/llms.txt) for the site index. Markdown versions are available by appending `.md` to content URLs or sending `Accept: text/markdown`. URL: https://nakafa.com/en/subjects/physics/kinematics/parabolic-movement-analysis Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/material/lesson/physics/kinematics/parabolic-movement-analysis/en.mdx Learn how to calculate peak time, range, and instantaneous velocity in projectile motion from velocity components. --- ## Start by Splitting the Initial Velocity Projectile analysis almost always starts with one simple question: how much of the velocity is horizontal, and how much is vertical? If the initial speed is $$v_0$$ and the launch angle is $$\theta$$, the initial components are: Visible text: Projectile analysis almost always starts with one simple question: how much of the velocity is horizontal, and how much is vertical? If the initial speed is and the launch angle is , the initial components are: Component: MathContainer Children: ```math v_{0x}=v_0\cos\theta ``` ```math v_{0y}=v_0\sin\theta ``` After that, the horizontal and vertical directions are calculated with different rules. The horizontal direction uses uniform motion because there is no horizontal acceleration. The vertical direction uses uniformly accelerated motion because gravity keeps acting downward. > The key to projectile analysis is using the same time for two different directions. Component: ParabolicMovementAnalysisLab Props: - title: Cannonball Component Analysis - description: A cannonball crosses the water while the ghost balls mark equal time steps from the same formulas used in the calculation. - labels: { chooseScenario: "Choose launch", scenarioNames: { "sixty-degree": <>Sixty Degrees, "long-drive": <>Long Drive, "high-arc": <>High Arc, }, factLabels: { horizontalComponent: <>Horizontal component, verticalComponent: <>Vertical component, peakTime: <>Peak time, flightTime: <>Flight time, range: <>Range, instantaneousVelocity: <>Velocity at two seconds, }, viewLabel: "Cannonball projectile analysis visual", } The arc and ghost balls show positions at equal time intervals, not separate horizontal and vertical paths. The values below the scene come from the same split: $$v_{0x}$$ stays constant, while $$v_y=v_{0y}-gt$$ changes because gravity acts downward. Visible text: The arc and ghost balls show positions at equal time intervals, not separate horizontal and vertical paths. The values below the scene come from the same split: stays constant, while changes because gravity acts downward. The number flow is easier to read as one connected formula sequence: ```math \begin{aligned} v_{0x}&=v_0\cos\theta\\ v_{0y}&=v_0\sin\theta\\ t_{\text{peak}}&=\frac{v_{0y}}{g}\\ T&=2t_{\text{peak}}\\ R&=v_{0x}T\\ \vec{v}(2)&=\langle v_{0x},v_{0y}-g(2)\rangle,\quad g=10\text{ m/s}^2 \end{aligned} ``` So the values below the scene are not manually chosen numbers. They are calculated from the selected launch, then shown as components, time, range, and instantaneous velocity. ## The Peak Happens When Vertical Velocity Is Zero At the peak, the object is still moving horizontally, but its up-and-down motion pauses for an instant. So the peak condition is $$v_y=0$$. Visible text: At the peak, the object is still moving horizontally, but its up-and-down motion pauses for an instant. So the peak condition is . ```math v_y=v_{0y}-gt ``` When $$v_y=0$$, the time to the peak is: Visible text: When , the time to the peak is: ```math t_{\text{peak}}=\frac{v_{0y}}{g} ``` For an initial speed of $$40\text{ m/s}$$ at an angle of $$60^\circ$$, the initial vertical component is $$20\sqrt{3}\text{ m/s}$$. Before the peak it is positive, at the peak it becomes zero, and after the peak it is negative because the object is moving downward. Visible text: For an initial speed of at an angle of , the initial vertical component is . Before the peak it is positive, at the peak it becomes zero, and after the peak it is negative because the object is moving downward. ## Maximum Range Uses Total Time If the object launches from the ground and returns to the ground at the same height, the total time is twice the time to the peak. Component: MathContainer Children: ```math t_{\text{total}}=2t_{\text{peak}} ``` ```math R=v_{0x}t_{\text{total}} ``` This formula shows why range is not determined by angle alone. Range also depends on horizontal velocity and how long the object stays in the air. ## Instantaneous Velocity Comes from Two Components To find velocity at a specific time, calculate $$v_x$$ and $$v_y$$ at that time first. The horizontal component stays constant, while the vertical component changes. Visible text: To find velocity at a specific time, calculate and at that time first. The horizontal component stays constant, while the vertical component changes. Component: MathContainer Children: ```math v_x=v_{0x} ``` ```math v_y=v_{0y}-gt ``` ```math v=\sqrt{v_x^2+v_y^2} ``` The direction of velocity can also be read from the ratio of the vertical and horizontal components: ```math \tan\alpha=\frac{v_y}{v_x} ``` If $$v_y$$ is positive, the velocity points upward. If $$v_y$$ is negative, the velocity points downward. Visible text: If is positive, the velocity points upward. If is negative, the velocity points downward. ## A Sixty Degree Launch Sets Two Components A ball is launched with an initial speed of $$40\text{ m/s}$$ and an angle of $$60^\circ$$. Use $$g=10\text{ m/s}^2$$. Visible text: A ball is launched with an initial speed of and an angle of . Use . This example uses an angle with clear trigonometric components, so the full analysis stays visible: split the initial velocity, get time from vertical motion, then use the same time to read range and instantaneous velocity. The initial components are: Component: MathContainer Children: ```math v_{0x}=40\cos60^\circ=20\text{ m/s} ``` ```math v_{0y}=40\sin60^\circ=20\sqrt{3}\text{ m/s} ``` The time to the peak is: ```math t_{\text{peak}}=\frac{20\sqrt{3}}{10}=2\sqrt{3}\text{ s} ``` The total time and range are: Component: MathContainer Children: ```math t_{\text{total}}=4\sqrt{3}\text{ s} ``` ```math R=20(4\sqrt{3})=80\sqrt{3}\text{ m} ``` Suppose we want the velocity at $$t=2\text{ s}$$. The vertical component is: Visible text: Suppose we want the velocity at . The vertical component is: ```math v_y=20\sqrt{3}-10(2)=20\sqrt{3}-20\text{ m/s} ``` Meanwhile, $$v_x=20\text{ m/s}$$ stays constant. The speed is: Visible text: Meanwhile, stays constant. The speed is: ```math \begin{aligned} v&=\sqrt{20^2+(20\sqrt{3}-20)^2}\\ &=\sqrt{2000-800\sqrt{3}}\\ &\approx 24.8\text{ m/s} \end{aligned} ``` The direction of the velocity from the horizontal is: ```math \alpha=\tan^{-1}\left(\frac{20\sqrt{3}-20}{20}\right)\approx 36.2^\circ ``` So at $$t=2\text{ s}$$, the ball is still moving upward at an angle because $$v_y$$ is still positive. Visible text: So at , the ball is still moving upward at an angle because is still positive.