Nov 13, 2012
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This question relates to the chapter re: Translational Motion < Freely Falling Bodies

What is the approximate vertical launch speed for an object that is at the same height above the ground at 2.7 s as it is at 5.3 s?

I'm confused about this part of the solution where it says, " We can conclude that it reached its apex at 4 s." How do they figure this part out?
 

BerkReviewTeach

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The easiest way to figure this out is to draw a parabola and put marks of equal height on the left and right sides to visualize it. The bottom left is t = o and the bottom right is the Tfinal. Mark 2.7 on the left side and 5.3 on the right side (at equal heights). If you move up both sides a little, you'll find that a mark on the left at 3 seconds will be at the same height as a mark at 5 seconds on the right side. This means the top must be at 4 seconds.

If that doesn't work, then consider this. At 2.7 seconds, it must be traveling up and at 5.3 seconds it must be traveling down. Given that it is at the same height, it must have 2.7 seconds of fall time still left at its 5.3-second height. That makes the total flight time 5.3 + 2.7 = 8 seconds. If it starts at 0 and finishes at 8 seconds, then it's at its highest point at 4 seconds.

Lastly you could simply average 2.7 and 5.3 to find that it was halfway between those two times at 4 seconds, and halfway through a parabola is the top.

From there the easiest thing to do is use the BR shortcut where vfinal = 10t in freefall, so from the top until the ground (which takes 4 seconds total) you reach an impact speed of roughly 40 m/s. Because of the symmetry of the parabolic flight, if the final speed is 40 m/s, then the launch speed must also be 40 m/s.
 
OP
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Nov 13, 2012
116
6
Bay Area
Status
Pre-Medical
The easiest way to figure this out is to draw a parabola and put marks of equal height on the left and right sides to visualize it. The bottom left is t = o and the bottom right is the Tfinal. Mark 2.7 on the left side and 5.3 on the right side (at equal heights). If you move up both sides a little, you'll find that a mark on the left at 3 seconds will be at the same height as a mark at 5 seconds on the right side. This means the top must be at 4 seconds.

If that doesn't work, then consider this. At 2.7 seconds, it must be traveling up and at 5.3 seconds it must be traveling down. Given that it is at the same height, it must have 2.7 seconds of fall time still left at its 5.3-second height. That makes the total flight time 5.3 + 2.7 = 8 seconds. If it starts at 0 and finishes at 8 seconds, then it's at its highest point at 4 seconds.

Lastly you could simply average 2.7 and 5.3 to find that it was halfway between those two times at 4 seconds, and halfway through a parabola is the top.

From there the easiest thing to do is use the BR shortcut where vfinal = 10t in freefall, so from the top until the ground (which takes 4 seconds total) you reach an impact speed of roughly 40 m/s. Because of the symmetry of the parabolic flight, if the final speed is 40 m/s, then the launch speed must also be 40 m/s.
That was a wonderful explanation -- thank you!
 

BerkReviewTeach

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That was a wonderful explanation -- thank you!
Intuitive answers that you can draw out are almost always the best way to go on MCAT PS questions.
 
Jul 22, 2011
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Easiest way to do this:
2.7 + 5.3 = 8.
8/2 = 4.
yo/g = time at the apex.
yo/g = 4 => 4 x g = 40.
 

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^ ya pretty much my method too.
TBRteach was spot on in the explanation. He should write/edit the books too haha
 

BerkReviewTeach

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^ ya pretty much my method too.
TBRteach was spot on in the explanation. He should write/edit the books too haha
Thank you for your kind words.

I did some of the editing in the latest physics book, but being as polite as possible I found that the politics of editing were rather tiring. I always tell students not to overthink on the MCAT, and low and behold I was part of a group overthinking which questions to keep in the books, which to remove, and which new ones best fit the book's goals. I think I'm done with editing for a while.