A ball is thrown straight up with an initial velocity of 6.4 m/s. It travels for 0.64 seconds, and has a change of position of 2.05 meters.

Question

A ball is thrown straight up with an initial velocity of 6.4 m/s. It travels for 0.64 seconds, and has a change of position of 2.05 meters. What is the ball’s final velocity?

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Minh Khuê 4 years 2021-08-02T13:52:36+00:00 1 Answers 19 views 0

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    2021-08-02T13:53:51+00:00

    Answer:

    V = 0.9 m/s

    Explanation:

    The parameters given are:

    Initial velocity U = 6.4 m/s

    Time t = 0.64s

    Height h = 2.05 m

    To find the final velocity, let us use third equation of motion

    V^2 = U^2 – 2gH

    Since the ball is going upward, g will be negative

    Substitute all the parameters into the formula

    V^2 = 6.4^2 – 2 × 9.8 × 2.05

    V^2 = 40.96 – 40.18

    V^2 = 0.78

    V = sqrt( 0.78)

    V = 0.883 m/ s

    V = 0.9 m/ s approximately

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