Hi!

I've recently got this homework piece that i've been struggling with.

It looks like this:

 

What i've come up with so far is this:

 

is it correct to just replace my known values to these equations?

 

Any help will be appreciated, thanks!

Is there some reason form making life so difficult?

 

What is the only acceleration of the body and is that constant?

Since this is supposed to be too hard for us to figure out on our own, I looked the equations up on the internet (as my teacher asked me to do)

The only acceleration should be the gravity constant g and it should'nt matter if it's 9,81 or 10,00 m/s/s

Edited May 20, 201510 yr by IvarV

I am sorry, I apologise I was too hasty the acceleration is not constant.

 

In order to derive your formulae you need you to set up and solve a differential equation.

 

Do you know anything about differential equaions and their solution by the method of separation of variables?

A common assumption is that the resistance is proportional to the square of the velocity.

 

In this case the equation of downward motion is then

 

[math]mg - kv^{2} = m \frac{\mathrm{d} v}{\mathrm{d} t}[/math]

 

Separating the variables the equation becomes

 

[math]kdt = \frac{mdv}{a^{^{2}}-v^{2}}[/math]

 

Where a² = mg/k

 

Integration yields

 

[math]c + kt = \frac{m}{a}tanh^{-1}\frac{v}{a}[/math]

 

Can you put the values given to obtain the constants?

Edited May 20, 201510 yr by studiot

yes, the topic is differential equations, but it usually has been okay to solve the problem in other ways.

Differential equations are what i need though.

 

Since i need the equations for distance and velocity, can i just move the variables around so that it opens velocity and then distance?

 

Sorry for my somewhat faulty English

 

Edit: I will try to replace the given values to this equation when i get home, thanks

Edit2: can you explain where the tan function comes from?

Edited May 20, 201510 yr by IvarV

It is not the tangent, it is the hyperbolic tangent

 

[math]tanh\left ( x \right )= \frac{e^{2x}-1}{e^{2x}+1}[/math]

 

I need to go out now, but post again later when you need more.

is the v in the formula the initial velocity?

Edited May 20, 201510 yr by IvarV

No, v is the velocity at time any time t.

 

It is the general solution giving the velocity as a function of time or (nearly) the answer to the first part of your question.

It remains to substitute the boundary conditions to obtain values for the constants.

 

I gave the formula for the hyperbolic tangent before.

 

The inverse tangent is

 

[math]tanh^{-1}x = \frac{1}{2}ln\left ( \frac{1+x}{1-x} \right )[/math]

could you look at it this way?

 

1. you are told air resistance is proportional to velocity

[latex]a_{air} = - j \cdot v(t) +h[/latex]

where j and h are slope and intercept of a linear equation. We know that:

[latex]a_{air}(39)+g=0[/latex]

because it stops accelerating at v=39 - ie gravity balances air resistance. We also know that

[latex]a_{air}(0) = 0[/latex]

because there is no air resistance at zero velocity

we can thus solve for j and h

 

2. we have an equation for a_{air} in terms of v

[latex]a_{air} = - j \cdot v(t) +h[/latex]

and we also know that a_{total} is a_{air} +a_{gravity}

This is a first order differential equation

[latex]\frac{dv(t)}{dt}=a_{gravity}+a_{air}=a_{gravity}- j \cdot v(t) +h[/latex]

which we can solve to find v(t)

 

3. Adding v(t) from above to v(0) we get an equation for velocity of the object wrto time

 

4. Integrating this with respect to time we get the position of the object wrto time

 

I have not included the equations I get

a. because I am unsure of the methodology until Studiot comments on it

b. because it is homework