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Homework Help: Acceleration due to gravity

  1. Jan 24, 2004 #1
    a 2.50-kg stone thrown upward from the ground at 14. m/s returns to the ground in 9. s

    my question is: is it possible to find the acceleration due to gravity? (keep in mind this is not taken from earth.)
  2. jcsd
  3. Jan 24, 2004 #2


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    Sure. Assuming that gravity is the only thing accelerating the object (i.e., there's no air resistance or anything), you can use the following equation:

    [tex] a = \frac{\Delta v}{\Delta t} [/tex]

    Here, vo = +14 m/s (the velocity it was thrown at, Δt = 9 s, the elapsed time, and v = -14 m/s. That initial kinetic energy was traded off for gravitational potential energy on the way up, and then given back as kinetic energy on the way down, so it ended up with the same magnitude velocity (remember that we said no energy was lost to air resistance or anything like that).
  4. Jan 24, 2004 #3
    Yes. The stone will have the same velocity when it hits the ground as it did when it originally left. Conservation of energy. As the stone reaches maximum height, all of its kinetic energy will have been converted to gravitational potential energy. As the stone falls, this potential energy will be converted back to kinetic energy.

    Knowing the original velocity and Time, acceleration is easy to calculate.

    lol, you beat me to it James!
  5. Jan 25, 2004 #4
    so is the acceleration 28/9 m/s^2?
  6. Jan 25, 2004 #5
    I missed something here. The original question stated: a 2.50-kg stone thrown upward from the ground at 14. m/s returns to the ground in 9. s

    This means that it took 9 seconds from the time it left the ground until it fell back down. that's 4.5 seconds each way. When the stone reaches it's maximum height, it will have a velocity of 0 m/s. We know that it will impact the ground at a V of 14 m/s and it will take 4.5 seconds to go from 0 m/s to 14 m/s. So...
    [tex] a = \frac{V}{t} = \frac{14 m/s}{4.5 s} = 3.11 m/s^2[/tex]
    Last edited: Jan 25, 2004
  7. Jan 25, 2004 #6
    Actually that works out as well since the round trip took nine seconds and +14 m/s - -14 m/s = 28 m/s. So 28/9 is correct as well.

    ugh, it's to late for this. :smile:
  8. Jan 25, 2004 #7
    Not to nitpick, but isn't "same speed" more appropriate than "same velocity"?
  9. Jan 26, 2004 #8
    Please, nitpick. I goofed. Yes, same speed would be more appropriate.
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