The video transcript discusses a physics problem involving projectile motion (gerak parabola) to calculate the final velocity of a projectile after a specific time.

Projectile Motion Calculation (Gerak Parabola)
📌 Initial conditions given: Initial velocity ($V_0$) = 60 m/s, elevation angle ($\theta$) = 30 degrees, gravity ($g$) = 10 m/s², and time ($t$) = 2 seconds.
🎯 The goal is to find the resultant velocity ($V$) at $t=2$ seconds using the formula $V = \sqrt{V_x^2 + V_y^2}$.
📏 Horizontal velocity component ($V_x$) remains constant: $V_x = V_0 \cos\theta = 60 \times \cos(30^\circ) = 60 \times \frac{1}{2}\sqrt{3} = \mathbf{30\sqrt{3}}$ m/s.

Vertical Velocity Component Analysis
⬆️ Initial vertical velocity ($V_{0y}$) calculation: $V_{0y} = V_0 \sin\theta = 60 \times \sin(30^\circ) = 60 \times \frac{1}{2} = \mathbf{30}$ m/s.
📉 Vertical velocity ($V_y$) at time $t$ is calculated using $V_y = V_{0y} - gt$.
🔧 Substituting values: $V_y = 30 - (10 \times 2) = 30 - 20 = \mathbf{10}$ m/s.

Final Velocity Determination
📐 The final velocity ($V$) squared is calculated: $V^2 = V_x^2 + V_y^2 = (30\sqrt{3})^2 + (10)^2$.
🧮 Calculation yields: $V^2 = (900 \times 3) + 100 = 2,700 + 100 = \mathbf{2,800}$.
✅ The final velocity is $V = \sqrt{2,800}$, which simplifies to $V = \sqrt{400 \times 7} = \mathbf{20\sqrt{7}}$ m/s.

Key Points & Insights
➡️ In projectile motion, the horizontal velocity ($V_x$) is constant as there is no horizontal acceleration.
➡️ The vertical velocity ($V_y$) is dependent on time ($t$) and gravity ($g$), requiring the use of kinematic equations for the Y-axis.
➡️ To find the resultant velocity ($V$) at any time, calculate the independent X and Y components first, and then use the Pythagorean theorem ($\sqrt{V_x^2 + V_y^2}$).

📸 Video summarized with SummaryTube.com on Oct 11, 2025, 06:22 UTC