Horizontally launched projectile | Two-dimensional motion | Physics | Khan Academy

Horizontally Launched Projectile Definition and Context
📌 A horizontally launched projectile is an object starting with zero initial vertical velocity and only a horizontal component to its launch velocity.
🧱 Examples include a ball rolling off a flat table or a person running straight off a cliff.
🧗 In the example provided, a diver is launched horizontally at $v_{0x} = 5 \text{ m/s}$ from a cliff $30 \text{ m}$ high ($\Delta y = -30 \text{ m}$ assuming downward is negative).

Decomposition of Motion
▶️ Horizontal ($x$) motion and vertical ($y$) motion occur independently of each other.
↔️ The horizontal velocity ($v_x$) remains constant at $5 \text{ m/s}$ throughout the flight (assuming no air resistance, $a_x = 0$).
⏬ The vertical motion is driven by gravity, starting with $v_{0y} = 0$ and accelerating downwards at $a_y = -9.8 \text{ m/s}^2$.

Solving for Time (Vertical Analysis)
🛑 The common trap is incorrectly setting the initial vertical velocity ($v_{0y}$) to the given horizontal velocity ($5 \text{ m/s}$); it must be zero.
⏳ Time ($t$) is found using the vertical displacement equation: $\Delta y = v_{0y} t + \frac{1}{2} a_y t^2$.
📉 Plugging in values: $-30 \text{ m} = (0)t + \frac{1}{2} (-9.8 \text{ m/s}^2) t^2$, yielding $t \approx 2.47 \text{ seconds}$.
⚠️ Forgetting the negative sign for displacement ($\Delta y = -30 \text{ m}$) results in attempting to take the square root of a negative number.

Solving for Horizontal Distance (Horizontal Analysis)
🎯 The calculated time ($t = 2.47 \text{ s}$) is the same for both dimensions, linking the vertical and horizontal analyses.
📏 The horizontal distance ($\Delta x$) is calculated using the simplified kinematic equation since $a_x = 0$: $\Delta x = v_{0x} t$.
🎯 Resulting horizontal distance: $\Delta x = (5 \text{ m/s}) \times (2.47 \text{ s}) \approx 12.4 \text{ meters}$.

Key Points & Insights
➡️ Identify the initial vertical velocity ($v_{0y}$) as zero for any object launched purely horizontally; this is the most frequent mistake in these problems.
➡️ Maintain coordinate convention: If downward is negative, the vertical displacement ($\Delta y$) must be negative (e.g., $-30 \text{ m}$).
➡️ Time is the linking variable: Solve for time ($t$) using the known variables in the vertical direction before solving for the unknown horizontal distance ($\Delta x$).

📸 Video summarized with SummaryTube.com on Jan 15, 2026, 16:42 UTC