EK physics elevator question (#293)

[Jwinsler7]

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The man in the elevator is holding the box that has a mass of 10kg. The floors of the building in which the elevator operates are separated by 5m. The elevator accelerate and decelerates at 1m/s^2, and moves at a maximum velocity of 5m/s.

Q. On a trip from the 1st to the 6th floor, the man wondered if he could throw the box into the air the moment the elevator leaves the 1st floor, and catch it the moment the elevator arrives at the 6th floor. What would the initial vertical velocity of the box have to be in order for this to work?

A) 5 m/s
B) 10 m/s
C) 37 m/s
D) 55 m/s

Here's my approach.
The distance from the 1st to the 6th floor is 25m, so that means the distance for the box to travel from its initial position to the top of its projectile would be 12.5m since projectiles are symmetric.

Once in the air, acceleration (deceleration) would be all due to gravity, so -10m/s^2. When the box is on top, its velocity would be 0m/s.

So I used Vf^2=V1^2 + 2ax.
0=v1^2 + 2(-10)(12.5), then v1^2= 500, v1= ~22m/s. But that's not one of the answer choices. Answer is D... 55 m/s

Any flaw in my thinking? Let me know. Thank you

 

[V5RED]

If it started with an initial velocity of 22m/sec then in 10 seconds(when the guy is at the 6th floor) the box would have already been sitting on the floor of the elevator for a while.

This problem is ridiculous imo though because unless the elevator is roofless there is no way he could hurl a 10kg box at 55m/s(123miles per hour) in an elevator that can go no faster than 5m/s without hitting the roof. Also, unless this guy is the incredible Hulk, hurling a 10kg box up at 55m/s is freaking ridiculous especially when it will be falling at 50m/s when he is supposed to catch it.

 

[Jwinsler7]

How did you come up with 10 seconds?
Considering the half portion of the projectile, couldn't I simply use vf^2=v1^2+2(a)(x) since I already know vf=0, a=-10 and x is 12.5?

I want to understand why this equation doesn't give me the correct answer..

 

[Singhp03]

This problem is ridiculous imo though because unless the elevator is roofless there is no way he could hurl a 10kg box at 55m/s(123miles per hour) in an elevator that can go no faster than 5m/s without hitting the roof. Also, unless this guy is the incredible Hulk, hurling a 10kg box up at 55m/s is freaking ridiculous especially when it will be falling at 50m/s when he is supposed to catch it.

same thing i thought of when reading this lol

 

[V5RED]

How did you come up with 10 seconds?
Considering the half portion of the projectile, couldn't I simply use vf^2=v1^2+2(a)(x) since I already know vf=0, a=-10 and x is 12.5?

I want to understand why this equation doesn't give me the correct answer..

Vf is not 0. Vf is something like 50m/s downward(ie his hands should be broken catching the box) If the box if thrown such that it reaches the 6th floor with zero velocity, it will not be there at the same time as the magic man in the magical roofless elevator.

 

[shaboobly]

How did you come up with 10 seconds?
Considering the half portion of the projectile, couldn't I simply use vf^2=v1^2+2(a)(x) since I already know vf=0, a=-10 and x is 12.5?

I want to understand why this equation doesn't give me the correct answer..

i think it's just arbitrary. think about this problem in real life because after all, all these things pertain to a real situation. you can just use poe to find the right answer. at least that's what i would have done.

 

[Jwinsler7]

Vf is not 0. Vf is something like 50m/s downward(ie his hands should be broken catching the box) If the box if thrown such that it reaches the 6th floor with zero velocity, it will not be there at the same time as the magic man in the magical roofless elevator.

I am only considering the half of the projectile (when the box going up and comes to a stop right before it starts falling). When the box reaches the top of its projectile, right before it starts falling, its velocity would be 0m/s. I am using this as the final velocity since projectiles are always symmetric.

Distance to get to the top is 12.5m, knowing that it the distance up and distance down would be the same. Vf is 0m/s. Vi is what we are looking for. a is g.
I fail to see how v1 could be 55m/s using vf^2=vi^2+2ax

i think it's just arbitrary. think about this problem in real life because after all, all these things pertain to a real situation. you can just use poe to find the right answer. at least that's what i would have done.

How would you use POE for this problem?

 

[V5RED]

I am only considering the half of the projectile (when the box going up and comes to a stop right before it starts falling). When the box reaches the top of its projectile, right before it starts falling, its velocity would be 0m/s. I am using this as the final velocity since projectiles are always symmetric.

Distance to get to the top is 12.5m, knowing that it the distance up and distance down would be the same. Vf is 0m/s. Vi is what we are looking for. a is g.
I fail to see how v1 could be 55m/s using vf^2=vi^2+2ax

Projectiles are NOT always symmetric.

Factors such as wind resistance and differences between the height of launching and landing affect the possibility for symmetry. In this problem, the projectile starts on the 1st floor and ends on the 6th. This means the path cannot be symmetric because the launch and land points are not identical.

That said, if the projectile had a symmetrical path(which would be the case if it landed at the same height from which it was thrown) then its launch speed and its landing speed would be identical. In other words, vf would be the same as vi if the path was symmetrical.

 

[shaboobly]

How would you use POE for this problem?

there is a max velocity set for the elevator after it takes time for the elevator to accelerate to it. also there is no gravity slowing down the elevator. there is, however, gravity acting on the thrown object and it must be thrown really hard to over come gravity and be in the exact spot at the same time the elevator gets there.

as was mentioned, the elevator would need to not have a roof and the person must be the incredible hulk.

 

[Jwinsler7]

Projectiles are NOT always symmetric.

Factors such as wind resistance and differences between the height of launching and landing affect the possibility for symmetry. In this problem, the projectile starts on the 1st floor and ends on the 6th. This means the path cannot be symmetric because the launch and land points are not identical.

That said, if the projectile had a symmetrical path(which would be the case if it landed at the same height from which it was thrown) then its launch speed and its landing speed would be identical. In other words, vf would be the same as vi if the path was symmetrical.

Thanks for the clarification. For some reason I thought the path upwards would be identical to the path downwards.

there is a max velocity set for the elevator after it takes time for the elevator to accelerate to it. also there is no gravity slowing down the elevator. there is, however, gravity acting on the thrown object and it must be thrown really hard to over come gravity and be in the exact spot at the same time the elevator gets there.

Sorry, I don't quite follow how this could narrow down to choice D. Could you elaborate a little further?

 

[Jwinsler7]

I tried to solve it this way.

since the distance is 25m and v average of the elevator is 2.5m/s, time would be 10 seconds.

Using x=vi*t+(1/2)at^2, vi is 52.5m/s. I guess that's close to 55m/s?

I am wondering if there's a quicker way to solve this. Let me know

 

[stevenz]

Jwinsler has the right of it.

Basically, the velocity that you throw the box in order for it to JUST reach the 6th floor is NOT the same as the velocty you need to throw the box in order for you to be able to reach the 6th floor at the same time. In the former (what you have calculated), the box would not reach the 6th floor in time and would have landed on the floor by the time you reach the 6th floor.

I don't think there's really an easier way to solve the problem. If acceleration were constant I would suggest solving the problem from the frame of reference of the elevator, but in this case it's much easier to take the frame of reference of the Earth.

 

[shaboobly]

Thanks for the clarification. For some reason I thought the path upwards would be identical to the path downwards.

Sorry, I don't quite follow how this could narrow down to choice D. Could you elaborate a little further?

the velocity needs to be high enough so that the box remains in the air the entire time that the elevator is moving to the 6th floor, which means you need to through that box really hard, because gravity is acting on it, decelerating it. gravity is not acting to slow down the elevator however.

 

[Jwinsler7]

the velocity needs to be high enough so that the box remains in the air the entire time that the elevator is moving to the 6th floor, which means you need to through that box really hard, because gravity is acting on it, decelerating it. gravity is not acting to slow down the elevator however.

Yeah, I get the logic. But what I don't understand is just by using that logic, how do you get to choose 55m/s? For instance, why not 33m/s?