# Thread: Ball thrown vertically upwards

1. ## Ball thrown vertically upwards

if a ball is thrown vertically upwards with an initial speed of 20ms^-1 how do i show its maximum height and the time taken to reach this position in terms of g.

Also, if the origin is now taken to be the point of projection, and the unit vector i pointing vertically upwards, what will the force diagram look like under the quadratic model of air resistance?

I know that the acceleration at time t is a(t)i in which

a(t) = -g/b^2(v(t)^2+b^2), where v(t)i is the velocity of ball at time t, and b^2 = mg/0.2D^2.

However, how do i show this?

Cheers

2. Originally Posted by moolimanj
if a ball is thrown vertically upwards with an initial speed of 20ms^-1 how do i show its maximum height and the time taken to reach this position in terms of g.

Also, if the origin is now taken to be the point of projection, and the unit vector i pointing vertically upwards, what will the force diagram look like under the quadratic model of air resistance?

I know that the acceleration at time t is a(t)i in which

a(t) = -g/b^2(v(t)^2+b^2), where v(t)i is the velocity of ball at time t, and b^2 = mg/0.2D^2.

However, how do i show this?

Cheers
For the first part are you using air resistance or not?

The Free-Body-Diagram will have a downward force due to the objects weight and (on the way upward) will also have a downward force due to the air resistance.

For the rest, use Newton's 2nd:
$\displaystyle \sum F = -mg - 0.2 D^2v^2 = ma$

$\displaystyle a = -g - \frac{0.2 D^2}{m} v^2$

$\displaystyle a = -g - \frac{0.2 D^2g}{mg} \cdot v^2$

$\displaystyle a = -\frac{g}{\frac{mg}{0.2D^2}} \left ( v^2 + \frac{mg}{0.2D^2} \right )$

$\displaystyle a = -\frac{g}{b^2}(v^2 + b^2)$

-Dan

3. Thanks Topsquark

Now, if I write a=dv/dt, how do I determine the time t in terms of v,b,g, and v0, and also the time taken for the ball to reach its greatest height if m=0.02kg, D=0.03m, g=9.81ms^-2, and v0=20ms^-1

Do i just integrate the formula found for a(t)?

Thanks

4. By integrating I get v = gv^3/3b^2 - gv.

But how do i determine in terms of time t?

Do i have to differentiate again (dx/dt) to get a relationship to calculate time taken to reach its maximum height?

If so, I have differentiated and get x=-gv^4/12b^2 - gv^2/2.

Do i just substitute values given previously to get x?

I'm confused big time

5. Originally Posted by topsquark
$\displaystyle a = -\frac{g}{b^2}(v^2 + b^2)$
You tried to integrate over v, not t.

$\displaystyle \frac{dv}{dt} = -\frac{g}{b^2}(v^2 + b^2)$

$\displaystyle \int_{v_0}^v \frac{dv}{v^2 + b^2} = -\frac{g}{b^2}\int_0^t dt$

$\displaystyle \frac{1}{b} \left [ tan^{-1} \left ( \frac{v}{b} \right ) - tan^{-1} \left ( \frac{v_0}{b} \right ) \right ] = -\frac{gt}{b^2}$

So
$\displaystyle v = b ~ tan \left [ tan^{-1} \left ( \frac{v_0}{b} \right ) - \frac{gt}{b} \right ]$

Now you need to integrate that with respect to time again:
$\displaystyle \frac{dh}{dt} = b ~ tan \left [ tan^{-1} \left ( \frac{v_0}{b} \right ) - \frac{gt}{b} \right ]$

$\displaystyle h - h_0 = \int_0^t \left \{ b ~ tan \left [ tan^{-1} \left ( \frac{v_0}{b} \right ) - \frac{gt}{b} \right ] \right \} ~ dt$

$\displaystyle h = h_0 + \left ( \frac{b^2}{g} \right ) ln \left | \frac{cos \left ( \frac{gt}{b} - tan^{-1} \left ( \frac{v_0}{b} \right ) \right ) }{cos \left ( tan^{-1} \left ( \frac{v_0}{b} \right ) \right ) } \right |$

This can be simplified somewhat, but I'll leave it like it is.

You've got m=0.02kg, D=0.03m, g=9.81ms^-2, and v0=20ms^-1, so b = 10900 N/m^2.

To get the max height, then, solve
$\displaystyle v = 10900 ~ tan \left [ tan^{-1} \left ( \frac{20}{10900} \right ) - \frac{9.81t}{10900} \right ]$
for t when v = 0. (For reference, I get t = 0.20387337 s.)

Then plug that time into
$\displaystyle h = \left ( \frac{10900^2}{9.81} \right ) ln \left | \frac{cos \left ( \frac{9.81t}{10900} - tan^{-1} \left ( \frac{20}{10900} \right ) \right ) }{cos \left ( tan^{-1} \left ( \frac{20}{10900} \right ) \right ) } \right |$
(For reference, I get h = 2.03873294 m, or about 2.0 m.)

PLEASE double check my work for errors!

-Dan

6. Thanks again

But the equation looks too complex to me. if

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

The multiplying out we get

dv/dt= -gv^2/b^2 - g

I cant understand why the tan element was needed when it would be easier to just multiply out and then integrate. Can you please explain - i'm really confused. I can follow what you have done, but not why.

7. Originally Posted by moolimanj
Thanks again

But the equation looks too complex to me. if

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

The multiplying out we get

dv/dt= -gv^2/b^2 - g

I cant understand why the tan element was needed when it would be easier to just multiply out and then integrate. Mr F says: Integrate with respect to what? If you integrate with respect to t, the left hand side is v but the right hand side can't be done because it depends on v. If you integrate with respect to v, the right hand side is easily done byt the left hand side can't be done ..... integral of dv/dt wrt v does not give v!!!

Can you please explain - i'm really confused. I can follow what you have done, but not why.
The differential equation has the generic form $\displaystyle \frac{dv}{dt} = f(v)$. This type of differential equation has to be flipped before any integrating with respect to v can be done. This means that in your question the integration will therefore involve an arctan(v) .....

You have surely learned the technique for solving the generic differential equation $\displaystyle \frac{dy}{dx} = f(y)$ .....?

8. Thanks but I'm still lost and confused. Here is the actual question I am trying to answer just in case what I have wrote has come across wrong (i'm tryin to answer (iii):

9. Originally Posted by moolimanj
Thanks but I'm still lost and confused. [snip]
What do you not understand? You have to flip it before integrating! When you flip it, you have to integrate 1/(v^2 + b^2). This will give an arctan(v/b).

1. Have you been taught how to integrate DE's of the form $\displaystyle {\color{red}\frac{dv}{dt} = f(v)}$ ?

2. Do you know how to integrate 1/(v^2 + b^2) ?

If the answer to at least one of these questions is no, then that's where your real problem is.

10. Originally Posted by moolimanj
By integrating I get v = gv^3/3b^2 - gv.
$\displaystyle \int \frac{dv}{dt}~dt = v$
This is good.

$\displaystyle \int (v^2 + b^2) ~dt \neq \frac{1}{3}v^3 + b^2v$
because v = v(t). v is NOT the variable of integration!!

You are thinking of something like
$\displaystyle \int (t^2 + b^2)~dt = \frac{1}{3}t^3 + b^2t$

-Dan