• Sep 1st 2013, 09:40 AM
camorris
Right. It may be because I have hurt my back and am heavily medicated but I cannot do this. I have sat here for 4 hours and have started going insane. It is totally pathetic. What the hell am I doing wrong. It is ridiculous. I have looked for quadratic methods to what does not make any sense at all and am...wait nope... I have lost my mind.
• Sep 1st 2013, 04:08 PM
phys251
1. Go back to your Physics textbook and make sure you are clear on exactly what velocity and displacement mean, if you don't know their definitions by heart.
2. You need to visualize what's going on. If you and I stood by and observed this activity, what would we see? When would the ball be moving slowly? Fast?
3. Draw a diagram, filling in as much information as you know.
4. Go forth and solve.
• Sep 4th 2013, 07:52 AM
camorris

The initial velocity is 0 I think? would that be correct? Or would it be 72 kph (ie) 20 m/sec

I have defined velocity as Change in displacement(m) divided by time (m/sec)

I have defined displacement as Change of position (the formula quoted in the thumbnail above)
• Sep 4th 2013, 09:35 AM
camorris
I have found out the following things.

t = 125/20 = 6.25.

However, when I put this in the formula in my first post 's' does not equal 125?

Any ideas
• Sep 4th 2013, 01:47 PM
FelixFelicis28
Quote:

Originally Posted by camorris
I have found out the following things.

t = 125/20 = 6.25.

However, when I put this in the formula in my first post 's' does not equal 125?

Any ideas

You need to revisit your textbook - you cannot use $\text{time} = \frac{\text{distance}}{\text{speed}}$ because the ball is accelerating under gravity (i.e. its velocity is not constant which the equation assumes).

You are given that the distance $s$ travelled by the ball after $t$ seconds is:

$s = u_0 t + \tfrac{1}{2} g t^2$

a) This is somewhat ambiguous but I am going to interpret it as "the time for the ball to drop such that only a fifth of the height of the building remains to travel down" i.e. it has traveled 100m. Equate 100 to s to get:

$100 = 20t + \tfrac{1}{2} g t^2$ and solve the quadratic to get the time.

b) As above except the distance traveled by the ball this time is 125m once it reaches the ground.
• Sep 5th 2013, 10:07 AM
camorris
I'm usually not a quitter.

Which is a shame. You would not believe how much time I have put into trying to answer this and am no closer. I will look into quadratics again. I have found a book to use to try and get this.
• Sep 5th 2013, 10:11 AM
FelixFelicis28
Quote:

Originally Posted by camorris
I'm usually not a quitter.

Which is a shame. You would not believe how much time I have put into trying to answer this and am no closer. I will look into quadratics again. I have found a book to use to try and get this.

Ok, good luck.

Just for quick reference though, to solve a quadratic of the form $ax^2 + bx + c = 0 :$

If $ax^2 + bx + c = 0 \implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
• Sep 5th 2013, 12:55 PM
camorris
This is what I have so far. I am a little worried about trying this out now.

Does this make more sense as to where I am going wrong? And it still doesn't work.
• Sep 5th 2013, 01:10 PM
FelixFelicis28
Quote:

Originally Posted by camorris
This is what I have so far. I am a little worried about trying this out now.

Does this make more sense as to where I am going wrong? And it still doesn't work.

Almost but you've made a sign error. For question b) your equation is:

$125 = 20t + \tfrac{1}{2}gt^2 \implies \tfrac{1}{2}gt^2 + 20t - 125 = 0$

You've made a sign error on the $b^2 - 4ac$ part, you haven't taken into account the negative sign of 125.
• Sep 5th 2013, 01:37 PM
camorris
Well..good news.

I inputted the same equation with the minus for the 125 and to result of t once inputted into the original formula does not equal 125.
Thanks for your help by the way.

Can you spell this out for me now please?
• Sep 5th 2013, 02:19 PM
FelixFelicis28
Quote:

Originally Posted by camorris
Well..good news.

I inputted the same equation with the minus for the 125 and to result of t once inputted into the original formula does not equal 125.
Thanks for your help by the way.

Can you spell this out for me now please?

Hmm, it should have.

I'll work through part a) with you and then see if you can do part b).

As in my previous reply, I'm going to assume part a) means that the ball has traveled 100m down i.e. s = 100. So, we have:

$100 = 20t + \tfrac{1}{2}gt^2 \implies \tfrac{1}{2}gt^2 + 20t - 100 = 0 \implies gt^2 + 40t - 200 = 0$.

Thus:

$t = \frac{-40 + \sqrt{40^2 - 4\cdot g (-200)}}{2g} = 2 (\sqrt{6} - 1)$ if $g = 10 \text{ms}^{-2}$.

Substituting this back in:

$\tfrac{1}{2} g \Big( 2 (\sqrt{6} - 1)\Big)^2 + 20 \cdot \Big(2(\sqrt{6} - 1)\Big) = 100$ if $g = 10 \text{ms}^{-2}$

Think you can do the second one?

(we discard the negative solution for time).
• Sep 5th 2013, 08:53 PM
camorris
I should be able to now thank you.

Although why have you doubled the values. Ie 100 became 200. 20 became 40 and so on.

If I knew I had to do this then I may not have sat here like an idiot.

Thanks though I really appreciate it.
• Sep 6th 2013, 03:27 AM
FelixFelicis28
Quote:

Originally Posted by camorris
I should be able to now thank you.

Although why have you doubled the values. Ie 100 became 200. 20 became 40 and so on.

If I knew I had to do this then I may not have sat here like an idiot.

Thanks though I really appreciate it.

Just to make the LaTeX a little easier when I put it into the quadratic equation! It doesn't make any difference whatsoever, but look, if you have an equation:

$ax^2 + bx + c = 0 \iff 2ax^2 + 2bx + 2c = 0 \iff \alpha ax^2 + \alpha bx + \alpha c = 0$

if $\alpha$ is a constant.

In an equation, you can manipulate one side of it anyway you want to, as long as you balance it and do it on the other side as well i.e. I multiplied the entire left-hand side of my quadratic by 2 and did the same to the right-hand side but as the RHS is 0, 0 x 2 is still 0.
• Sep 8th 2013, 04:32 AM
camorris