Gerak Parabola, Menghitung Jarak Terjauh Saat Peluru Ditembakkan Dari Puncak Menara Membentuk Sudut

Calculation of Maximum Height Reached During Parabolic Arc ($Y_M$)
📐 The maximum height attained above the launch point is determined by:
$$Y_M = \frac{V_0^2 \sin^2(\theta)}{2g}$$
* Plugging in values: $Y_M = \frac{(20)^2 \cdot (\sin(30^\circ))^2}{2 \cdot 10} = 5 \text{ m}$.
* The Total Height ($H_{total}$) from the ground is the tower height plus this maximum rise: $H_{total} = 400 \text{ m} + 5 \text{ m} =$405 \text{ m}.

Analysis of Free Fall (Semi-Parabola Component)
⏱️ The time ($t$) required for the projectile to fall the total vertical distance ($H_{total} = 405 \text{ m}$) starting from the apex height is calculated using the free-fall equation (assuming initial vertical velocity at the apex of the *fall* segment is zero, or by using the total height and considering the launch velocity components):
* The vertical displacement equation used for the falling portion (from the highest point down to the ground) simplifies to $H_{total} = \frac{1}{2}gt^2$ if using the total vertical drop from the conceptual apex of the parabola's full range, but the speaker uses the total height $405 \text{ m}$ against the effective drop from the highest point, leading to: $405 = \frac{1}{2}(10)t^2 \implies 405 = 5t^2$.
* Solving for time: $t^2 = 81$, so the time of flight for this segment is $t = 9 \text{ seconds}$.
* The horizontal distance covered during this fall segment ($A$) is calculated using constant horizontal velocity ($V_{0x}$): $A = V_{0x} \cdot t$.
* $V_{0x} = V_0 \cos(\theta) = 20 \cdot \cos(30^\circ) = 20 \cdot \frac{1}{2}\sqrt{3} = 10\sqrt{3} \text{ m/s}$.
* Distance $A = (10\sqrt{3}) \cdot 9 =$90\sqrt{3} \text{ m}.

Total Horizontal Distance (Range)
🔗 The total horizontal distance ($X$) is the sum of the horizontal distance covered during the parabolic arc phase ($10\sqrt{3} \text{ m}$) and the horizontal distance covered during the falling phase ($90\sqrt{3} \text{ m}$).
* Total Range $X = 10\sqrt{3} \text{ m} + 90\sqrt{3} \text{ m} =$100\sqrt{3} \text{ m}.

Key Points & Insights
➡️ The problem requires decomposing the trajectory into the initial parabolic segment (reaching maximum height relative to launch) and the subsequent segment involving the fall from the tower height.
➡️ The horizontal component of velocity ($V_{0x} = V_0 \cos\theta$) remains constant throughout the entire flight.
➡️ The total time of flight is determined by analyzing the total vertical displacement ($405 \text{ m}$) using the kinematic equation for vertical motion under gravity.
➡️ The final range is the sum of the horizontal distance covered while rising/peaking ($10\sqrt{3} \text{ m}$) and the horizontal distance covered while falling ($90\sqrt{3} \text{ m}$), totaling $100\sqrt{3} \text{ m}$.

📸 Video summarized with SummaryTube.com on Oct 11, 2025, 07:46 UTC