What is the initial velocity ($V_0$) and firing angle ($\theta$)?

The initial velocity ($V_0$) is 20 m/s, and the angle ($\theta$) is 30 degrees relative to the horizontal.

What is the height of the tower ($H$)?

The height of the tower ($H$) is 400 m.

What value was assumed for the acceleration due to gravity ($g$)?

Since $g$ was not specified in the problem, it was assumed to be $10 \text{ m/s}^2$.

What was the maximum height reached by the projectile above the launch point? ($Y_M$)

The maximum height reached ($Y_M$) above the launch point was calculated to be 5 m, resulting in a total drop height of 405 m.

How long did it take for the projectile to hit the ground after being fired? ($t$)

The total time of flight ($t$) until impact was calculated to be 9 seconds.

What is the final answer for the total horizontal distance from the base of the tower?

The total horizontal distance traveled ($X$) is $100\sqrt{3}$ meters.

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

This video explains how to calculate the maximum horizontal distance (range) a projectile travels when launched from a height (a tower), which involves analyzing both projectile motion and free fall components.

Calculation of Maximum Horizontal Range in Parabolic Motion ( $X_{max\_para}$ )
📏 The horizontal distance covered during the parabolic arc (from launch point to the height where it would complete its symmetric parabola) is calculated using the formula for $R$ or $X_{max}$ :
$X_{max\_para} = \frac{V_0^2 \sin(2\theta)}{g}$
Given $V_0 = 20 \text{ m/s}$ , $\theta = 30^\circ$ , and assuming gravitational acceleration ($g$) is $10 \text{ m/s}^2$ **.
* The initial calculation yielded $X_{max\_para} = 20\sqrt{3} \text{ m}$ .
* This distance is split equally on either side of the trajectory's peak: $10\sqrt{3} \text{ m}$ on each side of the peak line.

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: $H_{total} = 400 \text{ m} + 5 \text{ m} =$ .

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 $A = (10\sqrt{3}) \cdot 9 =$ .

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 $X = 10\sqrt{3} \text{ m} + 90\sqrt{3} \text{ m} =$ .

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

Projectile Launch Parameters
📌 Initial Velocity ( $V_0$ ): $20 \text{ m/s}$ .
⛰️ Tower Height ($H$): $400 \text{ m}$ .
📐 Launch Angle ( $\theta$ ): $30^\circ$ relative to the horizontal.
❓ Goal: Determine the total horizontal distance ($X$) from the base of the tower when the projectile hits the ground.

Calculation of Maximum Horizontal Range in Parabolic Motion ( $X_{max\_para}$ )
📏 The horizontal distance covered during the parabolic arc (from launch point to the height where it would complete its symmetric parabola) is calculated using the formula for $R$ or $X_{max}$ :
$X_{max\_para} = \frac{V_0^2 \sin(2\theta)}{g}$
Given $V_0 = 20 \text{ m/s}$ , $\theta = 30^\circ$ , and assuming gravitational acceleration ($g$) is $10 \text{ m/s}^2$ **.
* The initial calculation yielded $X_{max\_para} = 20\sqrt{3} \text{ m}$ .
* This distance is split equally on either side of the trajectory's peak: $10\sqrt{3} \text{ m}$ on each side of the peak line.

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: $H_{total} = 400 \text{ m} + 5 \text{ m} =$ .

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 $A = (10\sqrt{3}) \cdot 9 =$ .

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 $X = 10\sqrt{3} \text{ m} + 90\sqrt{3} \text{ m} =$ .

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

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

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