Kinematics help for physics?

Question

The question goes like this:
A ball is thrown straight up. At 12m, the ball reaches a velocity of 15m/s. What is the time it takes for the ball to go from the top of its path, to the bottom in seconds.

The answer should be 2.19 seconds but i need to know how to get there. I can normally do this but working with a variable in the middle confuses me

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General Physics RvTDLR 4 years 1 Answer 618 views 0

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Answer ( 1 )

  1. As the ball rises 12 meters, its velocity decreases at the rate of 9.8 m/s each second. Let’s use the following equation to determine the initial velocity.

    vf^2 = vi^2 + 2 * a * d
    15^2 = vi^2 + 2 * -9.8 * 12
    vi^2 = 460.2
    vi = √460.2

    This is approximately 21.4 m/s. When the ball reaches its maximum height, its velocity will be 0 m/s. Let’s use the following equation to determine the time for this to happen.

    vf = vi – g * t
    0 = √460.2 – 9.8 * t
    t = √460.2 ÷ 9.8

    This is approximately 2.19 seconds

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