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  • Projectile Motion: Calculating Height with Doubled Initial Velocity
    Here's how to solve this problem using the principles of projectile motion:

    Understanding the Concepts

    * Projectile Motion: An object moving under the influence of gravity alone is in projectile motion.

    * Initial Velocity: The velocity at which the object is launched.

    * Height: The maximum vertical distance the object reaches.

    * Gravity: A constant acceleration acting downwards (approximately 9.8 m/s²)

    Applying the Concepts

    1. Kinematic Equation: We can use the following kinematic equation to relate the initial velocity, final velocity, acceleration, and displacement (height):

    v² = u² + 2as

    Where:

    * v = final velocity (0 m/s at the highest point)

    * u = initial velocity

    * a = acceleration due to gravity (-9.8 m/s²)

    * s = displacement (height, h)

    2. First Scenario (Initial Velocity v):

    * v = 0 (at the highest point)

    * u = v

    * a = -9.8 m/s²

    * s = h

    Plugging these values into the equation:

    0² = v² + 2(-9.8)h

    h = v² / (2 * 9.8)

    3. Second Scenario (Initial Velocity 2v):

    * v = 0 (at the highest point)

    * u = 2v

    * a = -9.8 m/s²

    * s = h' (the new height)

    Plugging these values into the equation:

    0² = (2v)² + 2(-9.8)h'

    h' = (4v²) / (2 * 9.8)

    4. Comparing Heights:

    * Notice that h' = 4h

    Conclusion

    If the initial velocity is doubled (from v to 2v), the maximum height attained by the object will be four times the original height (h).

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