ek 1001 physics #124

[typicalindian]

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I have major problems with kinematics for some reason out of all the physics topics this one is the hardest for me -_-. could someone please explain this problem?

"One man drops a rock from a 100m 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 which height do the rocks meet?"

If someone could do it step by step I'd appreciate it, the book explanation doesn't really help me out. Thanks!

edit: there is a little diagram as well that shows the rock being thrown from the ground reaching the top of the building at the peak of its trajectory

 

[MedPR]

I don't understand how you can do this problem without knowing the velocity of the rock being thrown upward.

 

[MT Headed]

You can figure out the velocity. Tye second rock moves fast enough to just barely reach the top of the building. It is moving as fast, initially, as the first rock will be when it hits the ground.

 

[MedPR]

You can figure out the velocity. Tye second rock moves fast enough to just barely reach the top of the building. It is moving as fast, initially, as the first rock will be when it hits the ground.

But wouldn't the point that the rocks meet vary by the initial velocity of the thrown rock? How do you know the rock is being thrown only fast enough to just barely reach the top?

 

[typicalindian]

sorry there is a little diagram as well that shows the rock being thrown from the ground reaching the top of the building at the peak of its trajectory

 

[Ssina]

I think it'll just make it's point if it was going fast enough to just reach the top, as MT mentioned...

rock falling: Vi=0 and Vf=sqr 2gH m/s
rock thrown: Vi=sqr 2gH and Vf=0 m/s

for OP, meaning you essentially have all PEi converted to KEf in the first case; while all KEi is converted to PEf in the second.

 

[typicalindian]

bump...still doing get it

 

[MT Headed]

This question wants you to utilize two tricks.

One, to recognize the symmetry of motion of the two boxes, and to realize that they are not asking for the halfway point of distance of travel, but rather the halfway point of the time of travel.

Two, to remember that when something travels from rest and accelerates at a uniform rate, it spends half of its time in the first 1/4 of travel, and the other half of its time in the final 3/4 of travel.

Combining these two insights, the intersection ought to be 75m above the ground.

There isn't enough time on the real mcat to solve these problems the long way. These types of problems largely test your instincts and shortcuts.

 

[MedPR]

I still don't understand how you can assume that the Vi of the ball being thrown is in any way related to the Vf of the ball being dropped.

I understand that Vf=sqrt 2gh for the ball being dropped, but how can you assume that the person throwing the ball from the ground will be throwing it at that velocity?

 

[MrNeuro]

This question wants you to utilize two tricks.

One, to recognize the symmetry of motion of the two boxes, and to realize that they are not asking for the halfway point of distance of travel, but rather the halfway point of the time of travel.

Two, to remember that when something travels from rest and accelerates at a uniform rate, it spends half of its time in the first 1/4 of travel, and the other half of its time in the final 3/4 of travel.

Combining these two insights, the intersection ought to be 75m above the ground.

There isn't enough time on the real mcat to solve these problems the long way. These types of problems largely test your instincts and shortcuts.

huhhh??? how'd u know that i had no idea that half the time an object travels is spent in the first quarter of its descent......

I still don't understand how you can assume that the Vi of the ball being thrown is in any way related to the Vf of the ball being dropped.

I understand that Vf=sqrt 2gh for the ball being dropped, but how can you assume that the person throwing the ball from the ground will be throwing it at that velocity?

i was with you on that until i looked at the problem it clearly shows the object being thrown up is a projectile motion and reaches its peak at 100 m. Hence, because you know Vi, displacement, and acceleration (gravity) then you just solve using v2=u2 + 2as (s=displacement). which is vf=sqrt 2gh

 

[MedPR]

huhhh??? how'd u know that i had no idea that half the time an object travels is spent in the first quarter of its descent......



i was with you on that until i looked at the problem it clearly shows the object being thrown up is a projectile motion and reaches its peak at 100 m. Hence, because you know Vi, displacement, and acceleration (gravity) then you just solve using v2=u2 + 2as (s=displacement). which is vf=sqrt 2gh

Thanks, I'll take a look at the figure.

I think what MT headed is talking about has something to do with the equation h=1/2gt^2. But I'm not sure either..

 

[MT Headed]

Right. If an object falls at x=(some constant)t^2, and it takes two seconds to fall, where is it after the first second? It goes (some constant) x 1 meters.

How where is it after the second second? It is now at (some constant) x 4 meters.

So how far did it travel in the first second alone? one unit
How far did it travel in the last second alone? three more units, in order to reach a grand total of four units.

It spent half of its time going 1/4 the way.
It spent the other half of the time going the other 3/4 of the way.

 

[MedPR]

Right. If an object falls at x=(some constant)t^2, and it takes two seconds to fall, where is it after the first second? It goes (some constant) x 1 meters.

How where is it after the second second? It is now at (some constant) x 4 meters.

So how far did it travel in the first second alone? one unit
How far did it travel in the last second alone? three more units, in order to reach a grand total of four units.

It spent half of its time going 1/4 the way.
It spent the other half of the time going the other 3/4 of the way.

What?

In 1 second it goes 1 unit
2 seconds 4 units
3 seconds 9 units
4 seconds 16 units

first half is 5 units, second half is 25 units for a total of 30 units.

5/30 isn't 1/4

 

[Pisiform]

it's another exercise in h = h0 + v0t + (1/2)at2. Since we will need the initial velocity of the thrown rock, realize that the thrown rock must have an upwards initial velocity equal to the velocity a dropped rock would have just before it hits the ground (with the opposite sign). This initial speed of the thrown rock comes from v2 = 0 + 2(9.8)(100) = 1960 m/s2

Here's the method. Write down the height above the ground for the dropped rock, and for the height of the rock above the ground for the thrown rock. When they meet, these are the same. This will let you find the time they're at the same height, and that time will let you find the height.

So: the height of the dropped rock is

h = 100 + 0 + (1/2)(-9.8)t^2

The height of the thrown rock is

h = 0 + (sqrt1960)t + (1/2)(-9.8)t^2

When the rocks meet, h_dropped = h_thrown:

100 + 0 + (1/2)(-9.8)t^2 = 0 + (sqrt1960)t + (1/2)(-9.8)^t2

in which you can see the t^2 terms cancel, leaving you to solve for t:

100 = (sqrt1960)t

t =100/sqrt1960

Now take your pick of which equation to use for the height. Sub this value of t into it, and there's
your height when they meet.

hdropped = 100 - (1/2)(9.8)(100/sqrt1960)^2

h = 100 - 25 = 75 m :

hope that made some sense ....

 

[MT Headed]

What?

In 1 second it goes 1 unit
2 seconds 4 units
3 seconds 9 units
4 seconds 16 units

first half is 5 units, second half is 25 units for a total of 30 units.

5/30 isn't 1/4

It didn't travel 16 units during that last second, it traveled TO a displacement 16 units from the original position.

How far did it travel during the duration from 0 to 2 seconds in your example?
How much farther must it have traveled during the next 2 seconds, to get it to position X=16?

 

[MedPR]

It didn't travel 16 units during that last second, it traveled TO a displacement 16 units from the original position.

How far did it travel during the duration from 0 to 2 seconds in your example?
How much farther must it have traveled during the next 2 seconds, to get it to position X=16?

Oh right 🙂

So:

1 second = 1 unit
2 seconds = 1-4 units = 3 units
3 seconds = 4-9 units = 5 units
4 seconds = 9-16 units = 7 units.

First half = (3+1)/16 = 4/16 = 1/4
Second half = (5+7)/16 = 12/16 = 3/4

Wow, that's really cool 🙂 Thank you!

 

[MrNeuro]

WWWWWoooooooWWWWWWWW thanks!!! thats crazy

 

[typicalindian]

MT headed you are a physics genius! thanks a lot, you too pisiform the long way helped a lot 🙂

 

[Jay2910]

wait so how did you guys get sqrt 2gh? Yes, I get the point where you use v(f)^2=v(i)^2+2ax . . but I thought that the final velocity for the thrown rock is 0 at its apex . . .so if you say
0=v(i)^2+2ax . . .it doesn't make any sense anymore . .why? Or am I not supposed to be making that kind of assumption in the first place?

 

[Ssina]

from energy, at the bottom we have all KE and top it's all converted to PE, hence v=0m/s

 

[Jay2910]

Right now, is that v(f) or v(i)?

 

[FishyTheFish]

This question wants you to utilize two tricks.

One, to recognize the symmetry of motion of the two boxes, and to realize that they are not asking for the halfway point of distance of travel, but rather the halfway point of the time of travel.

Two, to remember that when something travels from rest and accelerates at a uniform rate, it spends half of its time in the first 1/4 of travel, and the other half of its time in the final 3/4 of travel.

Combining these two insights, the intersection ought to be 75m above the ground.

There isn't enough time on the real mcat to solve these problems the long way. These types of problems largely test your instincts and shortcuts.

This is a really nice solution, but how do you know that the point where they intersect occurs at the halfway point of the time of travel?

 

[ljc]

This is a really nice solution, but how do you know that the point where they intersect occurs at the halfway point of the time of travel?

Because they are both going the same distance, and have opposite velocities.

 

[FishyTheFish]

Because they are both going the same distance, and have opposite velocities.

If they meet at 75 meters above the ground, they don't go the same distance. The one dropped from the top travels 50 meters less. Right?

 

[ljc]

If they meet at 75 meters above the ground, they don't go the same distance. The one dropped from the top travels 50 meters less. Right?

Same total distance, sorry. They are both traveling 100m from start to finish in the same total time.

Maybe think of it this way: Object A @ point A is traveling to point B in 10s. Object B @ point B is traveling to point A in 10s. Given either constant acceleration or constant velocity, they will cross at 5s - there is no other possibility. I can't really think of a clear reason as to why, it's just something you'll need to think about for yourself.

 

[MedPR]

This question wants you to utilize two tricks.

One, to recognize the symmetry of motion of the two boxes, and to realize that they are not asking for the halfway point of distance of travel, but rather the halfway point of the time of travel.

Two, to remember that when something travels from rest and accelerates at a uniform rate, it spends half of its time in the first 1/4 of travel, and the other half of its time in the final 3/4 of travel.

Combining these two insights, the intersection ought to be 75m above the ground.

There isn't enough time on the real mcat to solve these problems the long way. These types of problems largely test your instincts and shortcuts.

Should be 25m above the ground right?

 

[MT Headed]

Should be 25m above the ground right?

The block at the top of the building starts from rest. In half the time, it will only travel 25m. The block at the bottom starts out launched at great speed so it can eventually reach the top of the building. It isn't starting from rest.

 

[MrNeuro]

it's another exercise in h = h0 + v0t + (1/2)at2. Since we will need the initial velocity of the thrown rock, realize that the thrown rock must have an upwards initial velocity equal to the velocity a dropped rock would have just before it hits the ground (with the opposite sign). This initial speed of the thrown rock comes from v2 = 0 + 2(9.8)(100) = 1960 m/s2

Here's the method. Write down the height above the ground for the dropped rock, and for the height of the rock above the ground for the thrown rock. When they meet, these are the same. This will let you find the time they're at the same height, and that time will let you find the height.

So: the height of the dropped rock is

h = 100 + 0 + (1/2)(-9.8)t^2

The height of the thrown rock is

h = 0 + (sqrt1960)t + (1/2)(-9.8)t^2

When the rocks meet, h_dropped = h_thrown:

100 + 0 + (1/2)(-9.8)t^2 = 0 + (sqrt1960)t + (1/2)(-9.8)^t2

in which you can see the t^2 terms cancel, leaving you to solve for t:

100 = (sqrt1960)t

t =100/sqrt1960

Now take your pick of which equation to use for the height. Sub this value of t into it, and there's
your height when they meet.

hdropped = 100 - (1/2)(9.8)(100/sqrt1960)^2

h = 100 - 25 = 75 m :

hope that made some sense ....

so would setting the time equal to each other work for all instances of figuring out where any two objects w/ varying velocities meet?

This question wants you to utilize two tricks.

One, to recognize the symmetry of motion of the two boxes, and to realize that they are not asking for the halfway point of distance of travel, but rather the halfway point of the time of travel.

Two, to remember that when something travels from rest and accelerates at a uniform rate, it spends half of its time in the first 1/4 of travel, and the other half of its time in the final 3/4 of travel.

Combining these two insights, the intersection ought to be 75m above the ground.

There isn't enough time on the real mcat to solve these problems the long way. These types of problems largely test your instincts and shortcuts.

so would setting the time equal to each other would work with any velocity or only with objects of the same velocity and same displacement?

 

[MT Headed]

Well, if the 'event' is defined as two objects meeting, then that implies that their displacements and their times have to be equal for the event, yes? Assuming both displacements and both times are measured from the same origins, of course.

 

[SaintJude]

Just to make sure this is clear to me. Is this following statement true ?

By the time both of these objects meet, the ball dropped from the top of the building would already have been through half of its motion time. And the ball thrown from the ground up would have gone through 3/4 of its motion time.

 

[MT Headed]

No, it also would have gone through half of its motion time. The bottom ball would have gone through 3/4 of its motion distance, but only half its motion time.

 

[MedPR]

it's another exercise in h = h0 + v0t + (1/2)at2. Since we will need the initial velocity of the thrown rock, realize that the thrown rock must have an upwards initial velocity equal to the velocity a dropped rock would have just before it hits the ground (with the opposite sign). This initial speed of the thrown rock comes from v2 = 0 + 2(9.8)(100) = 1960 m/s2

Here's the method. Write down the height above the ground for the dropped rock, and for the height of the rock above the ground for the thrown rock. When they meet, these are the same. This will let you find the time they're at the same height, and that time will let you find the height.

So: the height of the dropped rock is

h = 100 + 0 + (1/2)(-9.8)t^2

The height of the thrown rock is

h = 0 + (sqrt1960)t + (1/2)(-9.8)t^2

When the rocks meet, h_dropped = h_thrown:

100 + 0 + (1/2)(-9.8)t^2 = 0 + (sqrt1960)t + (1/2)(-9.8)^t2

in which you can see the t^2 terms cancel, leaving you to solve for t:

100 = (sqrt1960)t

t =100/sqrt1960

Now take your pick of which equation to use for the height. Sub this value of t into it, and there's
your height when they meet.

hdropped = 100 - (1/2)(9.8)(100/sqrt1960)^2

h = 100 - 25 = 75 m :

hope that made some sense ....

No, it also would have gone through half of its motion time. The bottom ball would have gone through 3/4 of its motion distance, but only half its motion time.

MT, can you point out where psiform went wrong (or where I'm misreading his math)?

 

[MT Headed]

He solved for h, the height in meters above the ground where the balls meet. Which is also the distance the bottom ball travels. We are all in violent agreement here.

 

[MedPR]

He solved for h, the height in meters above the ground where the balls meet. Which is also the distance the bottom ball travels. We are all in violent agreement here.

It looks like he solved for the distance the ball dropped.

"hdropped = 100 - .."

100-25 = ball dropped 75 meters.

Oh, wait nevermind. I see what I was doing now. Sorry

 

[Jay2910]

it's another exercise in h = h0 + v0t + (1/2)at2. Since we will need the initial velocity of the thrown rock, realize that the thrown rock must have an upwards initial velocity equal to the velocity a dropped rock would have just before it hits the ground (with the opposite sign). This initial speed of the thrown rock comes from v2 = 0 + 2(9.8)(100) = 1960 m/s2

Here's the method. Write down the height above the ground for the dropped rock, and for the height of the rock above the ground for the thrown rock. When they meet, these are the same. This will let you find the time they're at the same height, and that time will let you find the height.

So: the height of the dropped rock is

h = 100 + 0 + (1/2)(-9.8)t^2

The height of the thrown rock is

h = 0 + (sqrt1960)t + (1/2)(-9.8)t^2

When the rocks meet, h_dropped = h_thrown:

100 + 0 + (1/2)(-9.8)t^2 = 0 + (sqrt1960)t + (1/2)(-9.8)^t2

in which you can see the t^2 terms cancel, leaving you to solve for t:

100 = (sqrt1960)t

t =100/sqrt1960

Now take your pick of which equation to use for the height. Sub this value of t into it, and there's
your height when they meet.

hdropped = 100 - (1/2)(9.8)(100/sqrt1960)^2

h = 100 - 25 = 75 m :

hope that made some sense ....

This is great! Thanks!
I just would like to know one thing though . .here we are using 2 distance formulas, equating them to each other in order to get the time right. For the thrown rock . .you used v(f)=sqrt(1960) . .but in the distance equation that is substituted into the V(o) spot . . . .how did you know that it was going to work out?