Gerak Vertikal ke Atas Soal & Pembahasan Gerak Jatuh Bebas BELATIK fisika IPA SMA Peminatan

Summary

This video series discusses various aspects of motion, focusing on vertical motion. In part 7, the discussion centers on solving problems related to vertical projectile motion. Key topics include calculating the speed of a bullet 4 seconds after being fired, the speed at 47 meters above the ground, the bullet's height at different times, and its maximum height. The video also covers time calculations for reaching maximum height and falling back to the ground, providing a step-by-step breakdown of relevant equations and physics concepts.

Takeaways

• 📈 The video discusses key concepts of motion, including displacement, velocity, and speed, building on earlier videos.
• 🚀 In part 2, topics such as instantaneous velocity, average acceleration, and instantaneous acceleration were covered.
• 🏃‍♂️ Part 3 focuses on uniform linear motion, while part 4 moves into uniformly accelerated motion.
• 📝 Part 5 tackles problem-solving related to uniformly accelerated motion.
• 🎯 Part 6 covers vertical motion, including upward motion, downward motion, and free fall.
• 🔫 A bullet is fired vertically from a pistol at an initial velocity of 50 m/s from a height of 2 meters, with gravity approximated at 10 m/s².
• ⏳ The velocity of the bullet 4 seconds after being fired is calculated to be 10 m/s using the equation v = v₀ - g * t.
• 📏 The velocity of the bullet at 47 meters from the ground is calculated as 40 m/s, factoring in the initial height of 2 meters.
• 📐 The bullet's height from the ground 4 seconds after being fired is calculated to be 122 meters.
• 🏔 The maximum height reached by the bullet from the ground is 127 meters, using the formula H_max = v₀² / 2g.

Q & A

What topics were covered in the previous parts of the video series?

The previous parts covered various concepts: Part 1 discussed position, displacement, speed, and velocity; Part 2 covered instantaneous velocity, average acceleration, instantaneous acceleration, and deceleration; Part 3 focused on uniform linear motion; Part 4 explained uniformly accelerated linear motion; Part 5 involved exercises and solutions for uniformly accelerated motion; Part 6 covered vertical motion, including upward, downward, and free-fall movements.

What is the initial velocity of the bullet in the problem discussed in Part 7?

The initial velocity of the bullet is 50 m/s when it is fired vertically upwards.

How do you calculate the velocity of the bullet 4 seconds after it was fired?

The velocity after 4 seconds is calculated using the formula: velocity = initial velocity - (gravitational acceleration × time). Substituting the values: V = 50 m/s - (10 m/s² × 4 s) = 50 m/s - 40 m/s = 10 m/s.

What is the formula to calculate the velocity at a specific height from the ground?

The formula used is: v² = v₀² - 2 × g × h, where v is the final velocity, v₀ is the initial velocity, g is gravitational acceleration, and h is the height above the starting point.

What is the velocity of the bullet when it reaches a height of 47 meters above the ground?

Using the formula v² = v₀² - 2 × g × h, with v₀ = 50 m/s, g = 10 m/s², and h = 45 m (since the pistol's height is 2 meters), the velocity is 40 m/s.

How do you calculate the height of the bullet from the ground after 4 seconds?

The height is calculated using the formula: height = (initial velocity × time) - (1/2 × gravitational acceleration × time²). Substituting the values: height = (50 m/s × 4 s) - (1/2 × 10 m/s² × 16 s²) = 200 m - 80 m = 120 m. Adding the initial 2 meters, the total height is 122 meters.

What is the formula to calculate the maximum height reached by the bullet?

The formula for maximum height is: Hmax = (v₀²) / (2 × g), where v₀ is the initial velocity and g is the gravitational acceleration.

What is the maximum height reached by the bullet from the ground?

The maximum height is calculated as Hmax = (50 m/s)² / (2 × 10 m/s²) = 2500 / 20 = 125 meters. Adding the 2 meters from the pistol's height, the total maximum height is 127 meters.

How much time does it take for the bullet to reach the maximum height?

The time to reach maximum height is calculated using the formula: time = initial velocity / gravitational acceleration. Substituting the values: time = 50 m/s / 10 m/s² = 5 seconds.

What is the total time taken for the bullet to rise and fall back to the ground?

The total time is the sum of the time to reach maximum height (5 seconds) and the time for the free fall back to the ground. The free-fall time is calculated as: t = √(2 × Hmax / g), where Hmax is 127 meters. Substituting values: t = √(2 × 127 m / 10 m/s²) ≈ 5.04 seconds. The total time is approximately 10.04 seconds.