A stone is thrown vertically upward with a speed of 29.0 m/s and when it reaches a height of 13 m, the velocity is 24.2 m/s. Usi

Question

A stone is thrown vertically upward with a speed of 29.0 m/s and when it reaches a height of 13 m, the velocity is 24.2 m/s.

Using the formula x = v0 • t + ½ • a • t2, find the time it takes to reach this height? Why do you get two values for time? Explain.

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Gia Bảo 5 years 2021-08-13T17:34:45+00:00 1 Answers 20 views 0

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  1. nguyetanh
    0
    2021-08-13T17:36:28+00:00
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    Answer:

    the value of t = 0.49 seconds shows that its upward journey

    and

    at t = 5.43 seconds shows in downward journey

    Explanation:

    Given:

    initial speed, u = 29 m/s

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

    h = 13 m

    Let it is moving with velocity v at a height of 13 m.

    Use third equation of motion

    v² = u² + 2gh

    By substituting the values

    v² = 29² – (2 * 9.8 * 13)

    v = sqrt 585.94

    v  = 24.2 m/s

    Let it takes time t to reach at height 13 m

    Use second equation of motion

    s = u * t + 1/2 * g * t²

    13 = 29t – 4.9t²

    4.9t² – 29t + 13 = 0

    using quadratic equation to solve time

         29 ± \sqrt{29^2 - 4 * 4.9 * 13}\\

    t = ————————————

                         2 * 4.9

    t = 5.43 second or t = 0.49 second

    Therefore…

    the value of t = 0.49 seconds shows that its upward journey

    and

    at t = 5.43 seconds shows in downward journey

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