The h Bar

@S007, the text editor now reads latex in real time

@MaryStar Hi Mary. Your question is off topic here as we specifically prohibit homework questions.

@MaryStar, please see physicsproblems.nfshost.com

Anonymous

10:05 AM

@Mew @Mew LOL...site promotion ;-P

@MaryStar in general for equations like this you use one of the SUVAT equations.

Her question is more on topic on the physics problems site

In this case you want the equation v^2 = u^2 + 2as

@MaryStar, if you post the question here, physicsproblems.nfshost.com I will provide a comprehensive solution. This site isn't the place though

Secret

@Obliv ??? I haven't ping you recently, so perhaps I should be the one to ask that question

10:06 AM

Where the acceleration is just the gravitational acceleration g

Anonymous

@MaryStar Sorry I meant equations of motion as John Rennie pointed out. Try that out.

@Mew you're going to be accepting questions that show no effort?

@JohnRennie We have that a=g=1, u is the initial velocity and s the final position, or not?

s is the distance travelled.

@JohnRennie, no she will have to set out what she's attempted so far

10:10 AM

which in this case is the distance travelled downwards when the object hits the ground.

Anonymous

@MaryStar s is the height fallen i.e. 10

Ah ok... So, we have that the initial velocity is 10, therefore,
v^2 = u^2 + 2as => v^2 = 10^2 + 2*10 => v^2=100+20 => v^2=120 => v=\sqrt{120}
Is this correct? @S007 @JohnRennie

@MaryStar the question is a little more complicated than that because the initial velocity is sideways not vertical.

The initial velocity is given as 10i so it's moving in the positive x direction.

Ah yes, since it is 10i.
Is the initial vertical velocity equal to 0?

Anonymous

@MaryStar take initial velocity in y direction as 0

user116211

10:16 AM

@JohnRennie halp.

user116211

I need help in MathJax.

The vertical component of the velocity is initially zero, so in the equation you set u=0 and the value of v calculated will be the vertical velocity when the object hits the ground.

v^2 = u^2 + 2as => v^2 = 2*10 => v^2=20 => v=\sqrt{20} => v=2sqrt{5} @S007

Is this correct?

If we are ingoring atmopheric drag then the horizontal velocity is unchanged.

@MaryStar That is the correct result for the vertical component of the velocity.

Anonymous

@MaryStar Seems correct. Now finally square the velocity in horizontal direction and add it to the square of velocity in vertical. After that take square root

10:18 AM

But I assume the question wants the total velocity so you need to combine the vertical and horizontal velocities.

When it hints the ground does it have also horizantal velocity?
For the horizontal do we have that a=0 and u=10 ?
@S007 @JohnRennie

Anonymous

@MaryStar yes

@MaryStar Yes. The horizontal velocity remains unchanged during the fall.

So, v=sqrt{10^2+20}=sqrt{120}, right? @S007 @JohnRennie

@MaryStar Yes

10:22 AM

I understand! Thank you very much!! :-)
@S007 @JohnRennie

Anonymous

@Mew does the direct message work on your site ? did you get my test message ?

Ah yes I did recieve it

the answer to your question is yes

Anonymous

oh all well then :)

Arnold Neumaier

@heather This is from an answer in meta by a user with only 15 points - not at all representative for users of the site against which you are ranting!

When we have that the motion of a particle in polar coordinates is givwn by the realtions $r=\cosh (t) \ \ \theta=t$, do we have that the acceleration is equal to $r''=\cosh (t)$ ?

10:53 AM

Ah no, it is $a_r=0$ and $a_{\theta}=2\sinh t$ and so $a=2\sinh t$, right? @S007 @JohnRennie

Anonymous

@MaryStar Sorry I don't know much about polar coordinates..maybe John can answer

Anonymous

@Mew Can you activate latex in comments ?

@S007, yes i will implement in time

not my number 1 priority though

wait

latex already is activiated in comments

or did you do something special

Anonymous

yes...now it is working

Anonymous

thanks

10:57 AM

np

@S007 No problem... Aboout the other problem http://physicsproblems.nfshost.com/?qa=112/at-what-distance-will-the-particle-be
at the moment I cannot send my comment...
We have that $F=mv\frac{dv}{dx} \Rightarrow -kx^2=mv\frac{dv}{dx} $. We are looking what distance it has made for $t=2$ right? So we have to solve for $dx$, or what?

Anonymous

@MaryStar Form a differential equation. Do you know how to solve them ?

Anonymous

btw why can't you comment there ?

@S007 I don't know... I will try again...