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idku
 one year ago
Math  Based  Physics question.
idku
 one year ago
Math  Based  Physics question.

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idku
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441307711913:dwdw:1441308026115:dw

idku
 one year ago
Best ResponseYou've already chosen the best response.1`NOTE: in the corner of the picture 2, all it says is "1.5 m".`

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2well you now know that its horizontal velocity is 9m/s and that it travels a vertical distance of 5.5m before it stops rising that second piece of info allows you to calculate its vertical velocity as it leaves her hand use the equation of motion \(v^2 = u^2 + 2ax\) v = 0, a = g, x = 5.5m, so get u

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2you follow? or questions?

idku
 one year ago
Best ResponseYou've already chosen the best response.1First of, how do I know that it travles 5.5 meters before it stops rising?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2the question says it rises to 7m above ground, it start from 1.5m above ground, 7  1.5 = 5.5m

idku
 one year ago
Best ResponseYou've already chosen the best response.1Oh, ok, 5.5m is our vertical displacement. Good. And what does "u" stand for?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2u = initial velocity v = final velocity yet another notation but it's simple and really easy on the eye eg because no subscripts

idku
 one year ago
Best ResponseYou've already chosen the best response.1Oh, good. And v=0 because that is the point where it changes from positive (the height of the ball increasing) to negative (the height of the ball decreasing). Am I right?

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large v^2=u^2+2ax\) \(\large 0^2=u^2+2(9.81)(5.5)\) like this?

idku
 one year ago
Best ResponseYou've already chosen the best response.1or, am I plugging +9.81 ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2this is right \[\large 0^2=u^2+2(9.81)(5.5)\]

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large v^2=u^2+2ax\) \(\large 0^2=u^2+2(9.81)(5.5)\) \(\large 0^2=u^2107.91\) \(\large u=\sqrt{107.91}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1yes, if I plugged in +9.81 I would have got imaginary velocity.

idku
 one year ago
Best ResponseYou've already chosen the best response.1u=10.39 m/s (this is initial velocy)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2*yes, if I plugged in +9.81 I would have got imaginary velocity.* nice! i'm not doing the sums so i trust you are happy with your calculation?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2OK, really important now. right now, we are working in the vertical direction! yes?!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2so "u=10.39 m/s (this is initial VERTICAL velocy)" [assuming your calculator is working!]

idku
 one year ago
Best ResponseYou've already chosen the best response.1I used wolfram, so should work, but the practice thingy doesn't accept 10.39

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2we already know the HORIZONTAL velocity, which is constant yes?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2we have to combine vertical and horizontal using pythagoreas

idku
 one year ago
Best ResponseYou've already chosen the best response.1I am not given the angle, how would I tell horizontal velocty/

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2i told you it in my first post go take a look and see where it came from

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2in case you're not getting this i will repeat it:p we already know the HORIZONTAL velocity, which is constant yes? i told you it in my first post go take a look and see where it came from bandwidth is an issue for me :(

idku
 one year ago
Best ResponseYou've already chosen the best response.1I got disconnected due to lightning. My apologies, I am here....

idku
 one year ago
Best ResponseYou've already chosen the best response.1yes, I see so I got 13.75 m/s http://www.wolframalpha.com/input/?i=%E2%88%9A%289%C2%B2%2B10.39%C2%B2%29

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2no worries! hope everyone's OK.

idku
 one year ago
Best ResponseYou've already chosen the best response.1No one is outside:) My garage is inside the house, so it's alright.

idku
 one year ago
Best ResponseYou've already chosen the best response.1I wish I had you as my physics teacher, you made everything as easy as even a nonmath person such as myself got it!

idku
 one year ago
Best ResponseYou've already chosen the best response.1There is just one more question (for this part at least), if you don't mind: `How high above the ground will the ball be when it gets to Julie?`

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2excellent, one thing you **must** remember we do the horizontal axis and vertical differently no acceleration in horizontal, g acts in the vertical then combine through pythagoreas that applies throughout mechanics and kinematics...

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2well you know how long it will take to get to Julie, right

idku
 one year ago
Best ResponseYou've already chosen the best response.1ok, i will take a note of pythagorean as regards to velocities as well as distances....

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2brill! you know ho far apart they are? you know the horizontal velocity? and so?

idku
 one year ago
Best ResponseYou've already chosen the best response.1Horizontal velocity is 9m/s. And the distance between them would be?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2didn't you work that out in an earlier part of the question?

idku
 one year ago
Best ResponseYou've already chosen the best response.1I did the maximum height for [the part in] picture 1, and I did horizontal and vertical displacements for that (and this) part(s).

idku
 one year ago
Best ResponseYou've already chosen the best response.1but that is the velocity of the first through, not the distance. So that is not it.

idku
 one year ago
Best ResponseYou've already chosen the best response.19(the horiz. velocity of the first through))

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2let me work it out first, it might jog your memory just wait

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2OK from the very first question, we know that it was thrown at 11m/s at 44deg we can use equation ofmotion to get flight time \[x = ut + \frac{1}{2}a t^2 = t(u + \frac{1}{2}at)\] follow so far? we are working in the vertical direction now.

idku
 one year ago
Best ResponseYou've already chosen the best response.1oh, so the displacementfory formula and displacementforx formula are all same, besides that every y is now x instead?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2always in these x = distance [technically displacement but don't worry about that] u = initial velocity v = final velocity a = acceleration t = time always here we are tracking the ball from the first girl to the second, and we are looking at motion in the vertical direction, where gravity plays a role

idku
 one year ago
Best ResponseYou've already chosen the best response.1wait, you said: \(x = ut + \frac{1}{2}at^2\) so x in this case is the vertical distance I am solving for. `a = g = 9.81m/s` u = 9m/s ` t = ?` how do I figure out time?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2we need to back to the first question to get the distance between the girls, because we know that one threw it and one caught it. so we know it started t 1.5m above ground, and ended there when caught. we also know it was thrown at 11m/s at 44drg to horizontal.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2and to do that we need first to calculate how long the ball spent in the air travelling between the girls there's a lot in there so take your time

idku
 one year ago
Best ResponseYou've already chosen the best response.1While the distance is same, the times are different at different speeds. So we can't use the first question to calculate the time of second through. (That would have been rediculous if I have done so).

idku
 one year ago
Best ResponseYou've already chosen the best response.1But, distance I can use that same displacement formula, applying everything I know from picture 1?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2no we need to know how far the girls are apart we cannot do that from the information in the second part of the question we need to go right back to the first part to do that that is why i suspect that you have already calculated that distance

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(12.32 = 9t + \frac{1}{2}(9.81)t^2\) ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=2%2811+sin+44%29%2811+cos+44%29+%2F+9.81 correct

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2OK so back to this last bit and we need to recap, i think. we know that the second girl threw it with horizontal velocity = 9m/s  vertical velocity = 10.39 m/s yes now how long will it take to reach the first girl? in the horizontal direction....

idku
 one year ago
Best ResponseYou've already chosen the best response.1Oh, I am setting incorrect equation, because we are working the time based on the information we have from the first question, so u=11

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(12.32 = 11t + \frac{1}{2}(9.81)t^2\) But I still get imaginary solutions

idku
 one year ago
Best ResponseYou've already chosen the best response.1with +9.81 I get real solutions:)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2just stop for a minute, i think we are crossing posts:p we know the distance between them is 12.327m we know that the second girl threw it with horizontal velocity = 9m/s  vertical velocity = 10.39 m/s agreed?

idku
 one year ago
Best ResponseYou've already chosen the best response.1yes, we had all of this either as given or found that:)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2so now we need to know how long that will take, for a ball with horizontal velocity = 9m/s to travel the distance between the girls, which is 12.327m

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(12.327= 9t + \frac{1}{2}(9.81)t^2\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2there is no gravity in the horizontal direction:p

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2just constant velocity in horizontal direction

idku
 one year ago
Best ResponseYou've already chosen the best response.1what is my formula for x then?

idku
 one year ago
Best ResponseYou've already chosen the best response.1which formula do I use, I mean

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2oooHHHHH!!!! think!

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(x = ut + \frac{1}{2}t^2\) ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2the first one you learn!

idku
 one year ago
Best ResponseYou've already chosen the best response.1Distance = Velocity × Time ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\[x = ut + \frac{1}{2}(0)t^2 = ut\] or t = distance / velocity !!!!

idku
 one year ago
Best ResponseYou've already chosen the best response.1oh, for horizontal veloccity the acceleration is 0. distance = (initial velocity) × (time) > time = distance ÷ (initial velocity) time = 12.327m ÷ 9m/s (when dividing by m/s, you multiply times s/m, so meters cancel we get: (12.327/9)s )

idku
 one year ago
Best ResponseYou've already chosen the best response.1for whatever that quotient is

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2yes, 1.37s http://www.wolframalpha.com/input/?i=2%2811+sin+44%29%2811+cos+44%29+%2F+%289.81+*+9%29

idku
 one year ago
Best ResponseYou've already chosen the best response.1Now we need to apply t=1.37 (time length interval of the ball's second through) to find the height of the ball when it gets to julie  that is when it travels 12.327 meters.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2one final recap before we finish we know the distance between them is 12.327m [but we don't need that info anymore] we know that the second girl threw it with horizontal velocity = 9m/s  vertical velocity = 10.39 m/s we know that the journey in th e**horizontal** direction too 1.37s. we need to know what happens in the vertical direction at t = 1.37. at what height is the ball? agreed?

idku
 one year ago
Best ResponseYou've already chosen the best response.1yes, I undertand that! You are a wonderful person!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2so we need an equation of motion with acceleration in it because of gravity \[x = ut + \frac{1}{2}at^2\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2we want to know x. x is the displacement of the ball in the vertical direction a = 9.91 m/s/s t = 1.37 s u = 10.39 m/s

idku
 one year ago
Best ResponseYou've already chosen the best response.1x is the variable of interest. `initial velocity  u=10.39m/s (we are interested in vertical velocity)` time  t=1.37s (time is same for Vert. and Horiz. displacement of the ball) `acceleration [due to gravity]  a=(g)=(9.81)`m/s² \(x = 10.39m/s \times 1.37s + \frac{1}{2}(9.81m/s^2)(1.37s)^2\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1so in the first and second part we have meters left only. Just making sure t makes sense the way I do it, so that if I encounter unit confusion I know I have done something wrong. \(x = 10.39 \times 1.37 + \frac{1}{2}(9.81)(1.37)^2\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1and whatever that is equal to....

idku
 one year ago
Best ResponseYou've already chosen the best response.1Verif. http://www.wolframalpha.com/input/?i=12.327%3D9t%2B%281%2F2%29%289.81%29t%C2%B2

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2yes so she throws it 5.03m above the first girls hand. add 1.5m onto that if you want distance above ground

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2oh! good you got it yourself:p

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2well, that's the right "way" to do it, any mistakes are calculations.....we now know "how" to do it:p

idku
 one year ago
Best ResponseYou've already chosen the best response.1so the main things I need are: \(\large v^2=u^2+2ax\) \(\large x=ut+\dfrac{1}{2}at^2\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1and of course I know how to find horizontal and vertical velocity components based on a given angle and initial velocity....

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2let me type them out, there are 3 \[v = u + at\] \[x = ut + \frac{1}{2}at^2\] \[v^2 = u^2 + 2ax\] trick is to pick the right one to use as they all look at different things. eg the last one doesn't have t in it. but it is incredibly useful if you know of it.

idku
 one year ago
Best ResponseYou've already chosen the best response.1this is probably the most valuable post in my life. TY so much. No excuse me I got to go:) tnx for formula recap!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2yes  horizotal vs vertical there's an awful awful lot of info in there, you have done well! cheers!!

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(v = u + at\) \(v_x~~ =~~u~ \cos(\theta)\) \(v_y ~~=~~ u~\sin(\theta)\) \(d = ut + \frac{1}{2}at^2\) \(v^2 = u^2 + 2ax\) x = distance [technically displacement but don't worry about that] u = initial velocity v = final velocity a = acceleration t = time Ok, when I have to time to look more into it, i will. Thank you very much. I actually get this stuff now.
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