AP Physics 1: Kinematics 17: Projectile Part 2: Shot at an Angle (Symmetric)

Summary

This video explains the principles of projectile motion, focusing on a ball kicked at 20 m/s at a 37° angle. It breaks down the motion into horizontal and vertical components, calculating the maximum height (7.2 m), total time of flight (2.4 s), and horizontal distance traveled (38.4 m). It also discusses the velocity at maximum height (16 m/s) and the speed at landing (20 m/s). The analysis emphasizes the importance of understanding vector components and using trigonometric functions to solve projectile motion problems effectively.

Takeaways

• 😀 Projectiles launched at an angle follow a parabolic trajectory, with horizontal and vertical components of motion.
• 🏃‍♂️ The horizontal motion of a projectile is uniform, moving at a constant velocity throughout its flight.
• ⬆️ The vertical motion of a projectile is influenced by gravity, leading to a maximum height before it falls back down.
• 📏 The initial velocity of a projectile can be broken into horizontal (Vx) and vertical (Vy) components using trigonometric functions.
• 🔍 To find Vx, use the formula Vx = V * cos(θ); for Vy, use Vy = V * sin(θ).
• 🌄 For a projectile kicked at 20 m/s at a 37° angle, the horizontal component is approximately 16 m/s, and the vertical component is about 12 m/s.
• 🏔️ The maximum height (ΔY) can be calculated using the equation V^2 = V0^2 + 2aΔY, resulting in a height of 7.2 m for the given projectile.
• ⏳ The total time of flight for the projectile can be found by calculating the time to reach maximum height and doubling it, yielding 2.4 seconds.
• 📍 The horizontal distance (ΔX) traveled by the projectile can be calculated with ΔX = Vx * t, which results in a distance of 38.4 m.
• ⚡ The total speed at which the projectile lands can be determined by combining horizontal and vertical components, resulting in a final speed of 20 m/s.

Q & A

What is the trajectory of a projectile launched at an angle above the horizontal?

The trajectory is a parabolic path, where the projectile rises to a maximum height and then falls back to the same height from which it was launched.

How does the shadow of a projectile behave when light is shined from different angles?

When light is shined from above, the shadow moves at a constant velocity horizontally. When light is shined from the side, the shadow behaves like a falling object, first moving upwards, reaching a maximum height, and then coming back down.

What are the components of the initial velocity of a projectile?

The initial velocity can be broken down into two components: the horizontal component (vₓ) and the vertical component (vᵧ), determined using trigonometric functions based on the launch angle.

How do you calculate the horizontal and vertical components of the initial velocity?

The horizontal component is calculated using vₓ = v₀ * cos(θ) and the vertical component using vᵧ = v₀ * sin(θ), where v₀ is the initial speed and θ is the launch angle.

What formula is used to find the maximum height of a projectile?

The maximum height can be found using the formula v² = v₀² + 2aΔy, where v is the final vertical velocity (0 at maximum height), v₀ is the initial vertical velocity, a is the acceleration due to gravity, and Δy is the maximum height.

How long does it take for a projectile to reach its maximum height?

The time to reach maximum height can be calculated using the equation v = v₀ + at, solving for time (t) when the final velocity at maximum height is 0.

What is the relationship between the time of ascent and the total time of flight for a projectile?

The total time of flight is twice the time taken to reach maximum height, as the upward and downward journeys are symmetric.

How is the horizontal distance traveled by the projectile calculated?

The horizontal distance (range) is calculated using the formula Δx = vₓ * T, where T is the total time of flight and vₓ is the horizontal component of the velocity.

What speed does the projectile have just before landing?

The speed just before landing can be calculated using the Pythagorean theorem, combining the horizontal and vertical components of velocity at that moment.

What significance do the angles 37° and 53° have in projectile motion problems?

These angles correspond to a 3-4-5 triangle, allowing for simpler calculations of sine and cosine, which can lead to more manageable numbers in problem-solving.