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Free-Fall Motion Problem

  1. Oct 17, 2007 #1
    1. The problem statement, all variables and given/known data

    A research party on Jupiter in the year 3005 drops a steel ball from 300 meters above ground. It takes exactly 5 seconds to reach the ground.

    a. Calculate the value of g (the acceleration due to gravity) on the surface of Jupiter.
    b. What is the velocity of the steel ball the instant before it strikes the ground?

    2. Relevant equations
    .dont' know how to do it...please help

    3. The attempt at a solution

    None yet...dont' know how to do it...please help
  2. jcsd
  3. Oct 17, 2007 #2
    Start with
    [tex]a=\frac{\Delta v}{\Delta t}=\frac{\Delta v}{(\Delta t)^2}[/tex]
    Then do the problem like you would any such problem on Earth! ^_^
  4. Oct 17, 2007 #3
    Thanks...but i'm confused about the part that it give the distance=300 meters; 5 sec=the time. how do i get the change in velocity and the change in time?
  5. Dec 9, 2011 #4
    h = 300m
    t = 5s
    Vi = 0 m/s

    use the following equation: h= (Vi)(t) + (at2)/2
    substitute the given values: 300m = (0m/s)(5s) + [a(5s)2]/2
    then we get: 300m = 25s2a/2
    a = 24m/s2
  6. Dec 9, 2011 #5


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    Homework Helper

    Since you have already been given the answer - here is the conceptual way to answer.

    Basis: When a body undergoes constant acceleration from zero, the average velocity is one half of the final velocity.

    Here: 300m covered in 5 seconds → average velocity is 60 m/s

    Thus final velocity - just before hitting Jupiter is 120 m/s [part (b) done]

    In order to reach that velocity in 5 seconds, you must add 24 m/s each second, so the acceleration due to gravoity on Jupiter would appear to be 24 m/s2/. [part (a)]
    Last edited: Dec 9, 2011
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