Gerak Parabola • Part 3: Contoh Soal Gerak Parabola Dimulai dari Ketinggian Tertentu

Summary

This educational video from the Jendela Sains channel focuses on projectile motion, specifically analyzing a case where a projectile is launched from a height of 16 meters. The instructor breaks down complex concepts such as calculating velocity and position after one second, determining the time to reach the highest point, and finding the maximum height achieved by the projectile. Key formulas are presented, along with practical examples to clarify how to apply them in real scenarios. The lesson concludes with tips for solving similar problems in projectile motion, emphasizing the importance of understanding motion from a height.

Takeaways

• 😀 This video explains projectile motion, specifically parabolic motion, using a case study involving a projectile shot from a height.
• 😀 A projectile is launched from a cliff 16 meters high with an initial speed of 20 m/s at an angle of 53°.
• 😀 The first calculation involves determining the projectile's velocity and position after one second.
• 😀 The horizontal component of velocity is calculated as 12 m/s, while the vertical component after one second is found to be 6 m/s.
• 😀 The total velocity after one second is calculated to be approximately 6√5 m/s.
• 😀 The position of the projectile after one second is given as coordinates (12 m, 27 m).
• 😀 The video demonstrates how to calculate the time to reach the maximum height, which is found to be 1.6 seconds.
• 😀 The maximum height achieved by the projectile is calculated to be 28.8 meters above the ground.
• 😀 The total time of flight until the projectile hits the ground is determined to be 4 seconds.
• 😀 The horizontal distance traveled by the projectile upon landing is calculated as 48 meters.

Q & A

What is the main topic of the video?

The video focuses on understanding projectile motion, specifically parabola motion from a certain height, using a case study involving a bullet shot from a cliff.

What height is the bullet shot from?

The bullet is shot from a height of 16 meters above the ground.

What is the initial velocity of the bullet?

The bullet is shot with an initial velocity of 20 meters per second.

At what angle is the bullet shot?

The bullet is shot at an elevation angle of 53 degrees.

How do you calculate the horizontal component of the bullet's velocity?

The horizontal component of the velocity (Vx) is calculated using the formula V0 * cos(α), which gives 12 m/s for this scenario.

What is the vertical component of the bullet's velocity after one second?

The vertical component (Vy) after one second is calculated to be 6 m/s.

How do you determine the bullet's position after one second?

The position is determined using the equations for horizontal and vertical motion, resulting in coordinates (12 m, 27 m) after one second.

What is the time taken for the bullet to reach its maximum height?

The time taken to reach maximum height (t-max) is calculated to be 1.6 seconds.

What is the maximum height reached by the bullet?

The maximum height (Ymax) achieved by the bullet is 28.8 meters.

What is the total time the bullet remains in the air?

The total time the bullet is in the air before hitting the ground is calculated to be 4 seconds.

How do you calculate the horizontal distance traveled by the bullet when it hits the ground?

The horizontal distance is calculated using the horizontal velocity and total time in the air, resulting in 48 meters.