Air resistance/velocity problem

[Kawakaze]

Homework Statement

A ball is projected vertically upwards with an initial speed of 20ms^−1 at a
height of 1.5m above the ground. Model the ball as a sphere of diameter D, mass m, and
assume that the quadratic model of air resistance applies.

Show that the component of acceleration at time t in the upward
direction is given by
a(t) = − g/b^2 x (v^2+b^2)

where v is the speed of the ball at time t, and b2 = mg/0.2D2.

Homework Equations

The Attempt at a Solution

There are 2 forces acting on the sphere, its weight and the air resistance, both acting vertically downwards. I used Newtons 2nd law

ma = mg + 0.2D^2v^2

I figure that the question involves rearranging this expression in terms of a. Now I am stuck. As I can't rearrange this expression to get the one given in the question, this leads me to believe my approach is wrong

 

[PhanthomJay]

If you divide both sides of the equation by m to isolate a, after first substituting .2 d^2 =mg/b^2, as given, you've got it! The minus sign in the solution assumes downward as negative. You assumed downward as positive, which is fine.

 

[Kawakaze]

Thanks for the reply, but I still don't get how to do this. In particular where the extra b^2 comes from.

I rearranged the given expression for b^2 to get -

0.2D^2 = mg/b^2

After substituting in

ma = mg + (mg/b^2)v^2

Which I get to

a = g + (g/b^2)v^2

Which is close, but I see no point anywhere where an extra b^2 could appear from...

 

[PhanthomJay]

Thanks for the reply, but I still don't get how to do this. In particular where the extra b^2 comes from.

I rearranged the given expression for b^2 to get -

0.2D^2 = mg/b^2

After substituting in

ma = mg + (mg/b^2)v^2

Which I get to

a = g + (g/b^2)v^2

yes, now multiply the first term on the right of the equal sign by $b^2/b^2$, (which is 1, which doesn't change its value)
$a = g(b^2/b^2) + (g/b^2)v^2$, now factor:
$a = g/b^2( v^2 + b^2)$

 

[Kawakaze]

Thanks! But how did you know to do that? Is there a rule for it?

 

[PhanthomJay]

Thanks! But how did you know to do that? Is there a rule for it?

Gee, I don't know, I never did it this way before, I even surprised myself. If you hadn't provided the solution, I would have done it like

$a = g + (g/b^2)v^2$
$a = g(1 + (v^2/b^2))$
$a = g(1 + (v/b)^2)$

But given the solution, I sort of worked backwards.

 

[Kawakaze]

Job done, looks like we both picked up something new! :)

Thanks again!