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29 October, 09:57

A boy throws a rock with an initial velocity of 2.15 m/s at 30.0° above the horizontal. If air resistance is negligible, how long does it take for the rock to reach the maximum height of its trajectory?

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  1. 29 October, 11:16
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    t = 0.11 second

    The time taken to reach the maximum height of the trajectory is 0.11 second

    Explanation:

    Taking the vertical component of the initial velocity;

    Vy = Vsin∅

    Initial velocity V = 2.15 m/s

    Angle ∅ = 30°

    Vy = 2.15sin30 = 2.15 * 0.5

    Vy = 1.075 m/s

    The height of the rock at time t during the flight is;

    From the equation of motion;

    h (t) = Vy*t - 0.5gt^2

    g = acceleration due to gravity = 9.8m/s^2

    Substituting the given values;

    h (t) = 1.075t - 0.5 (9.8) t^2

    h (t) = 1.075t - 4.9t^2

    The rock is at maximum height when dh/dt = 0;

    dh (t) / dt = 1.075 - 9.8t = 0

    1.075 - 9.8t = 0

    9.8t = 1.075

    t = 1.075/9.8

    t = 0.109693877551 s

    t = 0.11 second

    The time taken to reach the maximum height of the trajectory is 0.11 second
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