A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the ball can re

Question

A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the ball can reach.

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niczorrrr 3 years 2021-09-04T16:05:39+00:00 1 Answers 2 views 0

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  1. Maris
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    2021-09-04T16:07:16+00:00
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    Answer:

    h = 16.9 m

    Explanation:

    When a ball is thrown upward, its velocity gradually decreases, until it stops for a moment, when it reaches the maximum height, while its height increases. Thus, the law conservation of energy states in this case, that:

    Kinetic Energy Lost by Ball = Potential Energy Gained by Ball

    (0.5)m(Vf² – Vi²) = mgh

    h = (0.5)(Vf² – Vi²)/g

    where,

    Vf = Final Speed of Ball = 0 m/s (Since, ball stops for a moment at highest point)

    Vi = Initial Speed of Ball = 18.2 m/s

    g = acceleration due to gravity = – 9.8 m/s² ( negative for upward motion)

    h = maximum height the ball can reach = ?

    Therefore, using values in the equation, we get:

    h = (0.5)[(0 m/s)² – (18.2 m/s)²]/(-9.8 m/s²)

    h = 16.9 m

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