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  • Calculating Velocity at the Highest Point of a Projectile Motion
    Here's how to solve this problem:

    Understanding the Physics

    * Free Fall: When a ball is thrown upwards, it's under the influence of gravity, which causes it to decelerate. This means its upward velocity decreases as it rises.

    * Symmetry: The motion of the ball going up and coming back down is symmetrical. This means the speed at which it's going upwards at a certain height is the same as the speed it's going downwards at that same height.

    * Velocity at the Highest Point: At the highest point, the ball's instantaneous velocity is zero.

    Solving the Problem

    1. Velocity at 2 seconds before the highest point: Since the motion is symmetrical, the velocity 2 seconds before the highest point is the same magnitude as the velocity 2 seconds after the highest point (when it's falling back down).

    2. Applying the Equation: We can use the following equation to find the velocity:

    * v = u + at

    Where:

    * v = final velocity (what we want to find)

    * u = initial velocity (0 m/s at the highest point)

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

    * t = time (2 seconds)

    3. Calculation:

    * v = 0 + (-9.8 m/s²) * 2 s

    * v = -19.6 m/s

    Answer: The velocity of the ball 2 seconds before it reaches the highest point is -19.6 m/s. The negative sign indicates that the velocity is directed downwards.

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