EK 1001 physics #124

[LoLCareerGoals]

One man drops a rock from 100m.
Another throw is up from the bottom (0m) such that it reaches 100m and then starts to drop.
At what height do they intersect?
This is one of the EK1001s so SN2ed asks you to do this in 30 sec

A. 25m
B. 50m
C. 75m
D. 80m

My calcs:

d1 = 100 - gt^2/2
d2 = Vot - gt^2/2

Vo is currently unknown, but Vo/g = tpeak, then:

d2 = 100 = Vo(Vo/g) - g*(Vo/g)^2/2
Vo^2/g - Vo^2/(2*g) = 100
Vo^2/(2*g) = 100
Vo^2 = 2000
Vo = 20*sqrt(5) = 20*2.4 = 48

then to find t @ which d1=d2:

100-gt^2/2 = 48*t - gt^2/2
48t = 100
t = 2
the d=100 - 10*4/2 = 80
D.
Which is incorrect.

Shorcuts? How in the f am I supposed to do this in 30 sec?

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[NuttyEngDude]

This is one of the EK1001s so SN2ed asks you to do this in 30 sec

I think it is ok if this takes you more than 30seconds. I am pretty sure his rules are meant to be guidelines and not hard and fast gospel.

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I got 80 first time i did it too. But if you realize that you are plugging back t^2 to find the distance, you can estimate the value better (comes out to 5 instead of 4).

So t is 10/sqrt(20) and guesstimating it i get ~2 for this. But if I realize T^2 is 100/20, that comes out to 5 exactly. so when you plug it back in, d = 100 - ((10*5)/2) = 75

there are probably better "shortcuts" but this is a good one for accuracy at least.

 

[Postictal Raiden]

This question definitely takes more than 30s to do. It took me about 2.5 mins to get the right answer.

What makes it even harder is that 80 and 75 are very close to one another, so you can't employ the POE to solve the problem. Otherwise, it's easy to eliminate A and B since you should intuitively know that the dropped rock is being accelerated from rest and the thrown rock had an initial velocity and undergoing deceleration, so the dropped rock will initially travel a shorter distance than the thrown one.

 

[LoLCareerGoals]

Yep. My error was in estimating sqrt(5). Stuff like that's been killing me, cosines/sines of standard angles too.

 

[mehc012]

Yep. My error was in estimating sqrt(5). Stuff like that's been killing me, cosines/sines of standard angles too.

Yeah, I ended up making a "Math shortcuts" Anki deck with log rules, first few prime lns, the quick estimation trick for log(3·10^x), the basic trig values, and all of the first square roots. It's actually sped up my calculator-free math considerably!

 

[sazerac]

You solve this quickly by recognizing that these rocks are following paths that are exactly opposite one another, and therefore they will meet at the halfway point of TIME in their journey. (10 seconds)

Every object that undergoes constant acceleration from rest (d=1/2 a t^2) will travel one unit of distance in one unit of time, and a grand total of four units of distance after two units of time. So the intersection will be at 100/4 = 25 meters from the top of the building. (10 seconds)

Next you have to figure out where the origin is in their answer choices. Looks like it is the ground, not the top of the building. (5 seconds)

Answer 75 meters, as measured from the ground. Elapsed time: 25 seconds.

 

[LoLCareerGoals]

You solve this quickly by recognizing that these rocks are following paths that are exactly opposite one another, and therefore they will meet at the halfway point of TIME in their journey. (10 seconds)

How did you know that? Both displacements are concave parabolas (due to -gt^2/2 terms), why would they intersect midway through the x-axis domain makes no intuitive sense to me.

Every object that undergoes constant acceleration from rest (d=1/2 a t^2) will travel one unit of distance in one unit of time, and a grand total of four units of distance after two units of time. So the intersection will be at 100/4 = 25 meters from the top of the building. (10 seconds)

I like this trick

 

[postbacpremed87]

How did you know that? Both displacements are concave parabolas (due to -gt^2/2 terms), why would they intersect midway through the x-axis domain makes no intuitive sense to me.


I like this trick

Straight from Berkeley Review!

 

[LoLCareerGoals]

I did do a derivation in full generality that does show:
t_intesection = 1/2 * t_peak

It took 1/2 of an A4 page. Hardly shorter than derivation in the OP.

I do understand the intuition of an accelerating body (from rest) covering 1/4 in t/2 and 4/4 in t.
Not so with deccelerating body from some Vo.
Edit:
Ok I think...I am starting to get there.
So we can do it like this:
In first half of deceleration, Vo goes from Vo to Vo/2 (since it drops linearly with time). Average speed for that period is (Vo+Vo/2)/2 = 3Vo/4
Average speed for that latter half is (Vo/2+0)/2 = Vo/4
We can use average speed now find that during the first half of the flight the 3/4 of h will be covered, which happens to be when 1/4 of h is covered by the falling object -> which is where they meet.

Think this was worth the effort. Thanks for pissing me off into getting this (no sarcasm here).
It is about getting used to using average speeds: area of the trapezoid = 1/2(h1+h2)*L