Projectile at an angle | Two-dimensional motion | Physics | Khan Academy

Khan Academy
16 Jun 201112:46
EducationalLearning
32 Likes 10 Comments

TLDRThe video script explains the physics of a rocket launching a projectile at a velocity of 10 m/s at a 30-degree angle above the horizontal. It details how to calculate the projectile's range by breaking down the velocity into horizontal and vertical components, using trigonometry and the principles of projectile motion. The calculation assumes negligible air resistance and results in a projectile range of approximately 8.83 meters, highlighting the application of mathematical concepts in real-world scenarios.

Takeaways
  • πŸš€ A rocket is launching a projectile at a velocity of 10 m/s at an angle of 30 degrees above the horizontal.
  • πŸ”„ The velocity of the projectile is broken down into horizontal and vertical components to simplify the problem.
  • πŸ“ The vertical component is calculated using the sine of the launch angle, resulting in 5 m/s.
  • πŸ•’ The time the projectile stays in the air is determined by the vertical component and acceleration due to gravity (g = -9.8 m/sΒ²).
  • πŸ”’ The time in the air is calculated to be approximately 1.02 seconds using the formula: Ξ”t = Ξ”v / g, where Ξ”v is the change in vertical velocity.
  • πŸƒ The horizontal component of the velocity is determined using the cosine of the launch angle, resulting in 5√3 m/s.
  • πŸ“ˆ The horizontal displacement is found by multiplying the horizontal velocity component by the time in the air, yielding approximately 8.83 meters.
  • πŸŒ™ The problem assumes negligible air resistance, simplifying the physics and allowing for the assumption of constant horizontal velocity.
  • πŸ“Š Trigonometry plays a crucial role in calculating both the vertical and horizontal components of the projectile's velocity.
  • πŸ›° The principles of projectile motion can be applied to understand the trajectory and distance of objects launched under similar conditions.
  • πŸ” The script provides a step-by-step approach to solving problems involving projectile motion, emphasizing the importance of breaking down complex motions into manageable components.
Q & A
  • What is the initial velocity of the projectile?

    -The initial velocity of the projectile is 10 meters per second.

  • What is the angle of the launch relative to the horizontal?

    -The angle of the launch is 30 degrees upwards from the horizontal.

  • How is the velocity vector broken down in the problem?

    -The velocity vector is broken down into its vertical and horizontal components to simplify the problem.

  • What is the method used to determine the vertical component of the velocity?

    -The vertical component is determined using the sine of the launch angle, which is 30 degrees in this case.

  • What is the value of the vertical component of the velocity?

    -The vertical component of the velocity is 5 meters per second, calculated as 10 meters per second (initial velocity) times the sin of 30 degrees (1/2).

  • How is the time the projectile stays in the air calculated?

    -The time the projectile stays in the air is calculated by dividing the change in vertical velocity by the acceleration due to gravity (-9.8 meters per second squared).

  • What is the final vertical velocity of the projectile when it returns to the ground?

    -The final vertical velocity of the projectile is -5 meters per second, indicating it is moving downward at the same magnitude but in the opposite direction of the initial velocity.

  • How is the horizontal component of the velocity determined?

    -The horizontal component is determined using the cosine of the launch angle, which is 30 degrees in this case.

  • What is the value of the horizontal component of the velocity?

    -The horizontal component of the velocity is 5√3 meters per second, calculated as 10 meters per second (initial velocity) times the cosine of 30 degrees (√3/2).

  • How is the total horizontal displacement of the projectile calculated?

    -The total horizontal displacement is calculated by multiplying the horizontal component of the velocity by the time the projectile is in the air (1.02 seconds).

  • What is the total horizontal distance the projectile travels?

    -The total horizontal distance the projectile travels is approximately 8.83 meters, calculated as 5√3 meters per second times 1.02 seconds.

  • Why is air resistance considered negligible in this problem?

    -Air resistance is considered negligible to simplify the problem and to allow for a more straightforward application of the principles of physics without the added complexity of drag forces.

Outlines
00:00
πŸš€ Rocket Launch and Velocity Vector Analysis

This paragraph introduces the scenario of a rocket launching a projectile at a velocity of 10 meters per second at an angle of 30 degrees above the horizontal. The main focus is on breaking down the velocity vector into its horizontal and vertical components to simplify the problem. The vertical component is used to calculate how long the projectile will be in the air, assuming negligible air resistance. The process involves using trigonometric functions (sine and cosine) to determine the components and apply the principles of physics to predict the projectile's trajectory.

05:00
πŸ“‰ Vertical Motion and Time Calculation

The second paragraph delves into the vertical motion of the projectile, considering the initial and final velocities and the effect of gravity. It explains how to calculate the time the projectile stays in the air using the change in vertical velocity and the acceleration due to gravity (9.8 m/sΒ²). The calculation leads to a result of 1.02 seconds, which is the time the projectile is in the air. This information is crucial for determining the horizontal displacement of the projectile.

10:00
πŸƒβ€β™‚οΈ Horizontal Displacement Calculation

The final paragraph focuses on calculating the horizontal displacement of the projectile. It uses the horizontal component of the velocity, found through the cosine of the launch angle (30 degrees), and multiplies it by the time the projectile is in the air (1.02 seconds) to find the total horizontal distance traveled. The horizontal component is determined to be 5√3 meters per second, and when multiplied by the time in the air, the resulting displacement is approximately 8.83 meters.

Mindmap
Keywords
πŸ’‘Rocket
A rocket is a vehicle that uses thrust from a rocket engine to propel itself, often used in space exploration or as a model in physics problems. In the video, a rocket is used to launch a projectile, and its velocity and direction are critical to understanding the problem of how far the projectile will travel.
πŸ’‘Projectile
A projectile is an object that is propelled through the air and is subject to the forces of gravity and air resistance. In the context of the video, the projectile is a rock-like object launched by the rocket, and its motion is analyzed to determine the distance it travels.
πŸ’‘Velocity
Velocity is a vector quantity that describes the speed and direction of an object's motion. It is fundamental in physics for analyzing motion and is used in the video to break down the motion of the projectile into its horizontal and vertical components.
πŸ’‘Direction
Direction refers to the path or line along which something moves or points. In the video, the direction is specified as 30 degrees upwards from the horizontal, which is crucial for determining the motion of the projectile.
πŸ’‘Components
In physics, components refer to the individual parts of a vector quantity, such as the horizontal and vertical parts of a velocity vector. The video breaks down the velocity of the projectile into its horizontal and vertical components to simplify the analysis of its motion.
πŸ’‘Trigonometry
Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of triangles, often used in physics to solve problems involving vectors. In the video, trigonometry is used to calculate the components of the projectile's velocity and to determine the distance it travels.
πŸ’‘Gravity
Gravity is the force that attracts two bodies with mass towards each other, and it is responsible for the motion of the projectile in the vertical direction. The video assumes negligible air resistance and considers gravity to determine how long the projectile stays in the air.
πŸ’‘Air Resistance
Air resistance is the force that opposes the motion of an object through the air. In the video, it is assumed to be negligible, simplifying the problem and allowing the horizontal velocity of the projectile to remain constant.
πŸ’‘Horizontal Component
The horizontal component is the part of a vector quantity that is directed along the horizontal axis. In the video, the horizontal component of the projectile's velocity is used to calculate the distance it travels in the absence of air resistance.
πŸ’‘Vertical Component
The vertical component is the part of a vector quantity that is directed along the vertical axis. In the video, the vertical component is used to understand the projectile's motion in the air and to calculate the time it spends in the air.
πŸ’‘Displacement
Displacement is the change in position of an object and is a vector quantity that considers both magnitude and direction. In the video, the horizontal displacement of the projectile is calculated by multiplying the horizontal component of its velocity by the time it is in the air.
Highlights

The rocket launches a projectile with a velocity of 10 meters per second at an angle of 30 degrees above the horizontal.

The problem is simplified by breaking down the velocity vector into its vertical and horizontal components.

The vertical component of the velocity is used to determine how long the rock stays in the air, independent of the horizontal component.

Air resistance is considered negligible in this problem, simplifying the physics of the motion.

The vertical component of the velocity is calculated using the sine of the launch angle, resulting in 5 meters per second.

The time the projectile stays in the air is determined by dividing the change in vertical velocity by the acceleration due to gravity.

The change in vertical velocity is 10 meters per second (from 5 m/s upwards to -5 m/s downwards).

The acceleration due to gravity used in the calculations is -9.8 meters per second squared.

The total time the projectile is in the air is calculated to be approximately 1.02 seconds.

The horizontal component of the velocity is found using the cosine of the launch angle, resulting in 5√3 meters per second.

The horizontal displacement is calculated by multiplying the horizontal velocity component by the time in the air.

The total horizontal displacement of the projectile is approximately 8.83 meters.

This analysis assumes constant horizontal velocity throughout the projectile's flight.

The method used can be applied to a variety of projectile motion problems, regardless of the initial conditions.

The next video will introduce an alternative method for solving the time of flight, applicable to more complex scenarios.

Transcripts
Rate This

5.0 / 5 (0 votes)

Thanks for rating: