aamc 6R ps question

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sotired

sotired
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another stupid kinematics equation tripped me up...

the question simply asked to tell what happens to the time t for a ball that was thrown vertically upward and came back down to the ground, when it is thrown on a planet where g/6.

anyway the explanation says the roundtrip time t is given by 2v/g = t. is this just a simple manipulation of v=at? and the 2 is because the time is to go up AND come back down?

your responses in this time of stress would be much appreciated 😳
 
You pretty much got it. Imagine a ball being thrown in the air at 10m/s. Assuming gravity to be 10m/s^2, it takes one second for the ball to go up and stop, and then one more second for the ball to come back down.

So 2(10)/(10) = 2 seconds, which is what logically makes sense.

If you started at 20 m/s, it would take two seconds for gravity to make it hit the apex, and then two more seconds to backtrack and go back down.

2(20)/(10) = 4 seconds round trip


Edit:

And if the gravitational force was, say, 5m/s^2 it would make the round trip double, since it would take 2 seconds to stop a 10m/s ball and then 2 more to bring it back down.
 
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