Equation for upward motion of the particle with air drag

[usha]

Homework Statement

A particle of mass m is thrown upward with velocity v and there is retarding air resistance proportional to the square of the velocity with proportionality constant k. if the particle attains a maximum height after time t, and g is the gravitational acceleration, what is the velocity v?

Homework Equations

The Attempt at a Solution

if i take D=kv^{2 and now two forces act on the particle F=mg and the retarding force but the options here are a) √(k/g)tan[√(g/k)t]
b)√(gk) tan [√(g/k)t]
c) √(g/k) tan[√(gk)t]
d) √(gk) tan[√(gk)t]
so i m having problem in figuring out how this tan factor come in ...}

 

[BiGyElLoWhAt]

Homework Statement

A particle of mass m is thrown upward with velocity v and there is retarding air resistance proportional to the square of the velocity with proportionality constant k. if the particle attains a maximum height after time t, and g is the gravitational acceleration, what is the velocity v?

Homework Equations

The Attempt at a Solution

if i take D=kv^{2 and now two forces act on the particle F=mg and the retarding force but the options here are a) √(k/g)tan[√(g/k)t]
b)√(gk) tan [√(g/k)t]
c) √(g/k) tan[√(gk)t]
d) √(gk) tan[√(gk)t]
so i m having problem in figuring out how this tan factor come in ...}

<sup>Let me just clean this up:

If I take ##D = kv^2 ## and now two forces act on the particle ##F = mg## and the retarding force (##D##?) but the options here are:
a) ##\sqrt{k/g}\ tan(\sqrt{g/k}\ t)##
b) ##\sqrt{gk}\ tan(\sqrt{g/k}\ t)##
c) ##\sqrt{g/k}\ tan(\sqrt{gk}\ t)##
a) ##\sqrt{gk}\ tan(\sqrt{gk}\ t)##

right?
Is the [edit]retarding[edit] force you're referring to ##D##?

Where did you get these options? Are these supposed to be final answers? Multiple choice question I'm assuming, no?</sup>

 

[haruspex]

If I take ##D = kv^2 ## and now two forces act on the particle ##F = mg## and the retarding force

So what equation do you get for the motion?

 

[AlephZero]

Did you make a typo in the answer options, or have you defined ##k## wrongly?

I don't think any of the options have the correct dimensions. For example the expressions inside ##\tan(\cdots)## are not dimensionless.

 

[haruspex]

Did you make a typo in the answer options, or have you defined ##k## wrongly?

I don't think any of the options have the correct dimensions. For example the expressions inside ##\tan(\cdots)## are not dimensionless.

How true... looks like the question should have said the drag force is mkv²; or maybe k in each option should be replaced by k/m.
Btw, if we assume that one of the options is then correct, the question can be answered merely by selecting the one with the right dimensions everywhere. There is no need to find or solve any equations of motion. Quick, but somewhat unsatisfactory, since we never get to discover how the tan function comes into it.

 

[AlephZero]

How true... looks like the question should have said the drag force is mkv²

That doesn't make much physical sense, because the drag force depends on the shape of the object but not on its mass.

Maybe the best option is wait for the OP to come back.

 

[haruspex]

That doesn't make much physical sense, because the drag force depends on the shape of the object but not on its mass.

I did not mean to suggest that such a k would have a physical meaning. I agree that the choice of k as the letter suggests the force should be kv², because k is commonly used that way in this context. But given that m is fixed here, there's nothing to stop the question setter from choosing to define k as force/(mv²).

 

[usha]

this question is from JEST 2013 (joint entrance screening test) and the answer is supposed to be (c) according to the answer key.

 

[haruspex]

this question is from JEST 2013 (joint entrance screening test) and the answer is supposed to be (c) according to the answer key.

If we ignore the 'missing mass' problem, and take the drag equation to be F = mkv², option (c) is the only one that's dimensionally correct. Do you see that?

 

[usha]

oh!.. yes never thought of taking F = mkv2... now i see it...thanks

 

[Magic Johnson]

My guess is
F_net = -F_weight - F_drag = m ⋅ dv/dt
-m ⋅ g - k ⋅ v² = m ⋅ dv/dt
dv/(k / m ⋅ v² + g) = -dt

Notice that
∫1/(1 + x²)dx = atan(x)
Then
v(t) = √( g ⋅ m / k ) ⋅ tan( -t ⋅ √( g ⋅ k / m ))
assuming initial velocity is 0

 

[haruspex]

My guess is
F_net = -F_weight - F_drag = m ⋅ dv/dt
-m ⋅ g - k ⋅ v² = m ⋅ dv/dt
dv/(k / m ⋅ v² + g) = -dt

Notice that
∫1/(1 + x²)dx = atan(x)
Then
v(t) = √( g ⋅ m / k ) ⋅ tan( -t ⋅ √( g ⋅ k / m ))
assuming initial velocity is 0

Yes, but the difficulty was that does not match any of the offered choices. To get one of those, you have to realize that k here need not be what you expect. You are only told it is a constant of proportionality, so it need not conform to the equation "drag=kv²".