physics task

[begadelavela]

Start speed of some object is 5 meter/s . Last 20 meters of fall object passes for 0.4 seconds.

What is the height from which object start to fall?

What is the finnal speed of falling?

Thanks guys I need this fast.

Sorry for my bad English.

[Brooksie359]

16 minutes ago, begadelavela said:

Start speed of some object is 5 meter/s . Last 20 meters of fall object passes for 0.4 seconds.

What is the height from which object start to fall?

What is the finnal speed of falling?

Thanks guys I need this fast.

Sorry for my bad English.

20=Vo×t+-.5×9.81×t^2

Solve for Vo which is the velocity at the beginning of the 20 meter part of the fall. from their plug Vo into V=Vo+-9.81t

That should give you the final velocity. also t is .4 seconds in this situation. as for getting the height of the fall you would need to find out how long it would take the ball to reach the final velocity for the initial velocity of 5m/s by solving Vfinal=5+-9.81t then plug that t into the equation 0=h+5t+-.5×9.81×t^2. that should give you your answer for h

[Dash Lambda]

2 hours ago, Brooksie359 said:

-snip-

For the second part, it would be easier just to use another UAM: V₁²=V₀²+2ad

Rather than solving for total time and then plugging that in to find distance, you can just solve directly for distance.

[Mug]

step 1: work out the speed in the last 20 metres of the fall:

v=s/t so 20/0.4 = 50 ms^-1

The question must assume this is terminal velocity (becase the velocity is constant) and that it is reached at 20m from ground.

 

step 2: find a suvat that works for you:

I've already done this for you and it's v²=u²+2as

Where v is end velocity, u is initial velocity, a is acceleration (due to gravity) and s is displacement

We're trying to work out s here (displacement) so then I got this formula: (v²-u²)/2a=s

Plug all the numbers in and you get (50²-5²)/2x9.81

Press equals and the answer is 126.15 m

 

step 3: that's not accounting for the 20 m at the end (only the distance it's accelerating for), so add on 20 m and you get:

146.15 m from the ground.

[Brooksie359]

4 hours ago, Dash Lambda said:

For the second part, it would be easier just to use another UAM: V₁²=V₀²+2ad

Rather than solving for total time and then plugging that in to find distance, you can just solve directly for distance.

Yes you could. me personally I don't like memorizing a bunch of equations so I only remember the particle displacement equation and the fact that most of the other particle equations are just derivatives of that. then again I'm pretty sure that equation is derived from the conservation of energy equation so I could see how using conservation of energy would be easier. 

[Brooksie359]

2 hours ago, Mug said:

step 1: work out the speed in the last 20 metres of the fall:

v=s/t so 20/0.4 = 50 ms^-1

The question must assume this is terminal velocity (becase the velocity is constant) and that it is reached at 20m from ground.

 

step 2: find a suvat that works for you:

I've already done this for you and it's v²=u²+2as

Where v is end velocity, u is initial velocity, a is acceleration (due to gravity) and s is displacement

We're trying to work out s here (displacement) so then I got this formula: (v²-u²)/2a=s

Plug all the numbers in and you get (50²-5²)/2x9.81

Press equals and the answer is 126.15 m

 

step 3: that's not accounting for the 20 m at the end (only the distance it's accelerating for), so add on 20 m and you get:

146.15 m from the ground.

It would state it has reached terminal velocity if that were the case. I've tutored many people in physics and really do they give a particle motion problem with terminal velocity unless they give drag force and want you to solve for the speed at which it reaches terminal velocity. the general thing with physics problems is that if you have to make assumptions to solve a problem the problem statement will list those assumptions. 

[Dash Lambda]

10 minutes ago, Brooksie359 said:

Yes you could. I personally don't like memorizing a bunch of equations so I only remember the particle displacement equation and the fact that most of the other particle equations are just derivatives of that. Then again, I'm pretty sure that equation is derived from the conservation of energy equation so I could see how using conservation of energy would be easier.

I never really memorized them explicitly, I just always solved the problem symbolically first and eventually started skipping bits I'd done enough times. It's always good practice to reduce the number-crunching needed.

Oh, and that one isn't derived from conservation of energy:

v = v₀+at  >>  t = (v-v₀)/a

d = v₀t+at²/2 = v₀((v-v₀)/a)+a((v-v₀)/a)²/2 = (v²-v₀²)/2a

v²-v₀² = 2ad  >>  v² = v₀²+2ad

[Brooksie359]

4 minutes ago, Dash Lambda said:

I never really memorized them explicitly, I just always solved the problem symbolically first and eventually started skipping bits I'd done enough times. It's always good practice to reduce the number-crunching needed.

Oh, and that one isn't derived from conservation of energy:

v = v₀+at  >>  t = (v-v₀)/a

d = v₀t+at²/2 = v₀((v-v₀)/a)+a((v-v₀)/a)²/2 = (v²-v₀²)/2a

v²-v₀² = 2ad  >>  v² = v₀²+2ad

You could also do the same thing from conservation of energy. f=ma

Work = f×d  

.5×m×v^2=.5×m×vo^2 + m×a×d

Cancel out the m and multiply by 2 and you have the same equation. 

[Mug]

48 minutes ago, Brooksie359 said:

It would state it has reached terminal velocity if that were the case. I've tutored many people in physics and really do they give a particle motion problem with terminal velocity unless they give drag force and want you to solve for the speed at which it reaches terminal velocity. the general thing with physics problems is that if you have to make assumptions to solve a problem the problem statement will list those assumptions. 

If this question was asked of me then this would be how I'd answer it, back in the day I did A-Level Physics and got a top grade in it, questions rarely gave you clues like this (so you had to assume so much) but then again, the guy didn't really say what level he was studying at. Since it didn't really have very much information, it's sensible to assume that it reached terminal velocity but now that I read your answer it seems more sensible.

[Dash Lambda]

2 minutes ago, Brooksie359 said:

You could also do the same thing from conservation of energy:

f = ma

Work = fd  

mv²/2 = mv₀²/2 + mad

Cancel out the m and multiply by 2 and you have the same equation. 

Ah... Well, I hadn't done it like that before, so I guess it's perfectly reasonable to approach it both ways.

I still need to spend a little more time on energy problems to start readily applying it like I do kinematics.

 

A nitpick I want to note, though, is that you shouldn't use an "x" to represent an inner product, especially when handling vectors.

[Brooksie359]

27 minutes ago, Dash Lambda said:

Ah... Well, I hadn't done it like that before, so I guess it's perfectly reasonable to approach it both ways.

I still need to spend a little more time on energy problems to start readily applying it like I do kinematics.

 

A nitpick I want to note, though, is that you shouldn't use an "x" to represent an inner product, especially when handling vectors.

Lol I'm typing this on my phone. I wasn't really trying to have perfect notation. just trying to get the message across as easily as possible.