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An object is thrown straight upward with an initial velocity of 1217 pixels per second. The vertical velocity of the object after 2 seconds is 0 due to gravity and the object is 1188 pixels above its starting point. The acceleration on the object due to gravity is 100 pixels per second squared. What is the mass of the object?

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    $\begingroup$ "pixels per second squared" is not a unit of force. $\endgroup$ Aug 20, 2013 at 23:05
  • $\begingroup$ @DanielLittlewood: Would it make sense to replace the word "pixels" with "meters"? $\endgroup$
    – Joncom
    Aug 20, 2013 at 23:07
  • $\begingroup$ No; in either case it would be a unit of acceleration. Do you perhaps mean the gravitational acceleration (usually called $g$)? $\endgroup$ Aug 20, 2013 at 23:08
  • $\begingroup$ My bad, yes, I mean acceleration. $\endgroup$
    – Joncom
    Aug 20, 2013 at 23:28

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Just like with real gravity, the acceleration and motion of an object is independent of its mass whether measured in meters or pixels. So you cannot determine its mass other than to say that it is not zero. In order to determine mass you need force.

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  • $\begingroup$ I didn't know that. Thank you. $\endgroup$
    – Joncom
    Aug 20, 2013 at 23:33
  • $\begingroup$ This is only an approximation (discovered by Galileo, and valid only when the object doesn't move too far and doesn't weigh too much), but it's the best you can do with the given information. Newtonian gravitationis a little more complex. $\endgroup$
    – dfeuer
    Aug 21, 2013 at 3:25
  • $\begingroup$ @dfeuer, I disagree. It is not an approximation, but an exact fact. Galileo stated the general observation, to be further confirmed by Newton and Einstein. The fundamental principal is that gravitational mass and inertial mass are exactly the same. The principal holds in for all masses (from a BB to a star) and for all distances traveled (from a dropped ball to an orbiting planet). $\endgroup$
    – Tpofofn
    Aug 22, 2013 at 21:04
  • $\begingroup$ @Tpofofn, here's a hint: Newton's second and third laws come into play as soon as you look beyond the instantaneous. $\endgroup$
    – dfeuer
    Aug 23, 2013 at 0:19
  • $\begingroup$ @dfeuer, I am not following you. $\endgroup$
    – Tpofofn
    Aug 23, 2013 at 0:24

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