Kinematics Mcat Problem

  • Thread starter mrlucky0
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  • #1
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One man drops a rock from a 100 m building. At exactly the same moment, a second man throws a rock from the bottom of the building to the top of the building. At what height do the rocks meet?

I have trouble understanding how there can even be an answer to this problem without more info:

Doesn't the initial vertical velocity of the thrown rock need to be known?! Surely the height at which the rocks meet will be different than if the thrown rock is thrown upwards at 100 m/s (lower) or 1000 m/s (higher) right? Am I to make an assumption here?
 

Answers and Replies

  • #2
Pengwuino
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Yes, you need to know the velocity that the person throws the rock up. However, you can ASSUME it's thrown with a velocity, v, and solve for the height that they meet, assuming they do meet.
 
  • #3
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Yes, you need to know the velocity that the person throws the rock up. However, you can ASSUME it's thrown with a velocity, v, and solve for the height that they meet, assuming they do meet.
Yes. In looking at the solution, they assumed that the rocks meet exactly halfway through their flight. Interestingly this is also the time when the thrown rock reaches it's max height.
 
  • #4
Pengwuino
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Hmm that's weird, there must be something missing in the question that would lead you to calculate that they would meet half way....
 
  • #5
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Hmm that's weird, there must be something missing in the question that would lead you to calculate that they would meet half way....
Oops, I meant to say, they assumed they meet halfway through their flight time. The result here is that for any height, the thrown rock always travels 3x more than the dropped rock, so in this case 75m and 25m respectively. Does that sound right?

Anyway, it's just one of the "combinations," 1/2*t and 1/2*t. The way I see it, there are endless combinations of flight times that would chance their meeting height, depending on the initial speed of thrown rock.

I just think this is a poor question.
 
  • #6
Lok
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There's nothing missing. The energy from the bottom-up rock should be exactly enough to propel it to that height. The up-bottom rock is or should not be thrown with a initial speed, just let go. In that case, if I interpreted the problem correctly (which is usually a more complicated problem than a physical one) there is only one solution.
 
  • #7
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There's nothing missing. The energy from the bottom-up rock should be exactly enough to propel it to that height. The up-bottom rock is or should not be thrown with a initial speed, just let go. In that case, if I interpreted the problem correctly (which is usually a more complicated problem than a physical one) there is only one solution.
Oh good point, I didn't think of that! You're saying that there is only one initial velocity for the bottom-up rock to be thrown such that it just reaches the top of the building.
 

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