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13 June, 10:31

Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 3.00 m above the ground and measure that it hits the ground 0.814 s later What is the acceleration of gravity near the surface of this planet?

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  1. 13 June, 14:26
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    Answer: The gravitational acceleration near the surface of this planet is 9.056 m/s^2

    Explanation:

    When you drop an object, the only acceleration acting on the object is the gravitational acceleration, so we can writhe the acceleration of the wrench as

    a (t) = - g

    where the minus sign is because the wrench will move downwards.

    for the velocity of the object, we integrate over time and get:

    v (t) = - g*t + v0

    where v0 is the initial velocity, in this case is 0, because the wrench is droped.

    integrating again, we will get the position of the wrench:

    p (t) = - (g/2) * t^2 + p0

    where p0 is the initial position, in this case is 3m above the ground.

    p (t) = - (g/2) * t^2 + 3m

    now, afther 0.814seconds the wrench hits the ground, so we have:

    p (0.814 s) = 0 = - (g/2) * (0.814s) ^2 + 3m

    now we need to solve this for g.

    - (g/2) * (0.814s) ^2 + 3m = 0

    (g/2) * (0.814s) ^2 = 3m

    (g/2) = 3m / * (0.814s) ^2 = 4.528m/s^2

    g = 2*4.528m/s^2 = 9.056 m/s^2
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