# veloctity of a ball.

• Feb 26th 2013, 08:26 AM
Tweety
veloctity of a ball.
Can anyone please guide/help me with this question, i have attached it to the post,
thank you.

The equation looks so complicated, I just dont know where to even begin?
• Feb 26th 2013, 11:18 AM
ILikeSerena
Re: veloctity of a ball.
Quote:

Originally Posted by Tweety
Can anyone please guide/help me with this question, i have attached it to the post,
thank you.

The equation looks so complicated, I just dont know where to even begin?

Hi Tweety! :)

Let's begin with $v=0$.
I believe you have an expression for $v$.
Can you substitute it?
And then rewrite it to find t?
• Feb 26th 2013, 11:16 PM
Tweety
Re: veloctity of a ball.
Thank you, I got $t= tan^{-1}\frac{\alpha V_{n}}{g}$ ?

Is that correct? Do i use this value to fine ymax
• Feb 27th 2013, 12:03 AM
ILikeSerena
Re: veloctity of a ball.
Quote:

Originally Posted by Tweety
Thank you, I got $t= tan^{-1}\frac{\alpha V_{n}}{g}$ ?

Is that correct? Do i use this value to fine ymax

You should have:

$t = \frac 1 {\alpha \sqrt g} \tan^{-1} \frac {\alpha v_n} {\sqrt g}$

This is the time that v will be zero, and y will be on its maximum.
The time t depends on the coefficient $\alpha$.

But yes, then you use this value to find $y_{max}$.
• Feb 27th 2013, 12:43 AM
Tweety
Re: veloctity of a ball.
Quote:

Originally Posted by ILikeSerena
You should have:

$t = \frac 1 {\alpha \sqrt g} \tan^{-1} \frac {\alpha v_n} {\sqrt g}$

This is the time that v will be zero, and y will be on its maximum.
The time t depends on the coefficient $\alpha$.

But yes, then you use this value to find $y_{max}$.

So I put this value into the second equation?

Which is very complicated, I dont know how to simplify it, I put t = ot the above , and got this

$y = \frac{1}{\alpha} ln(cos(tan^{-1}\frac{\alpha v_{n}}{\sqrt{g}} - \alpha\sqrt{g} \frac{1}{\alpha \sqrt{g}}) tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}$

I dont know how to simplify this
• Feb 27th 2013, 01:51 AM
ILikeSerena
Re: veloctity of a ball.
Quote:

Originally Posted by Tweety
So I put this value into the second equation?

Which is very complicated, I dont know how to simplify it, I put t = ot the above , and got this

$y = \frac{1}{\alpha} ln(cos(tan^{-1}\frac{\alpha v_{n}}{\sqrt{g}} - \alpha\sqrt{g} \frac{1}{\alpha \sqrt{g}}) tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}$

I dont know how to simplify this

It looks like you did not properly substitute.
When you properly substitute there is a lot that disappears because it's zero.
Btw, you could have simplified the fraction with $\alpha \sqrt g$ in it - it's just 1.
No need to do that now though, because you first need the proper substution.
Furthermore, if you can't reasonably simplify, that's fine. It means you are done and that you have the answer.
• Feb 27th 2013, 01:56 AM
Tweety
Re: veloctity of a ball.
Oh, okay

I this time I tried it i got,

$\frac{1}{cos(tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}} )$ is this correct?

Actually sorry missed out the ln term,

so I got $lncos(tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}} )$
• Feb 27th 2013, 02:42 AM
ILikeSerena
Re: veloctity of a ball.
Quote:

Originally Posted by Tweety
Oh, okay

I this time I tried it i got,

$\frac{1}{cos(tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}} )$ is this correct?

Actually sorry missed out the ln term,

so I got $lncos(tan^{-1}\frac{\alpha V_{n}}{\sqrt{g}}} )$

Much better!
But you still need a factor $\frac{-1}{\alpha^2}$ in front and then you have $y_{max}$.

Btw, if you want you can simplify more.
That is, you can use that $\cos(\tan^{-1}(\frac y x)) = \frac {x} {\sqrt{x^2+y^2}}$