Math - Based - Physics question.

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Math - Based - Physics question.

Mathematics
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`NOTE: in the corner of the picture 2, all it says is "1.5 m".`
well 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

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you follow? or questions?
First of, how do I know that it travles 5.5 meters before it stops rising?
the question says it rises to 7m above ground, it start from 1.5m above ground, 7 - 1.5 = 5.5m
Oh, ok, 5.5m is our vertical displacement. Good. And what does "u" stand for?
u = initial velocity v = final velocity yet another notation but it's simple and really easy on the eye eg because no subscripts
Oh, 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?
yes
\(\large v^2=u^2+2ax\) \(\large 0^2=u^2+2(-9.81)(5.5)\) like this?
or, am I plugging +9.81 ?
you got it right!
the first time?
this is right \[\large 0^2=u^2+2(-9.81)(5.5)\]
\(\large v^2=u^2+2ax\) \(\large 0^2=u^2+2(-9.81)(5.5)\) \(\large 0^2=u^2-107.91\) \(\large u=\sqrt{107.91}\)
yes, if I plugged in +9.81 I would have got imaginary velocity.
u=10.39 m/s (this is initial velocy)
*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?
OK, really important now. right now, we are working in the vertical direction! yes?!
so "u=10.39 m/s (this is initial VERTICAL velocy)" [assuming your calculator is working!]
I used wolfram, so should work, but the practice thingy doesn't accept 10.39
we already know the HORIZONTAL velocity, which is constant yes?
we have to combine vertical and horizontal using pythagoreas
I am not given the angle, how would I tell horizontal velocty/
?
Sorry.
i told you it in my first post go take a look and see where it came from
in 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 :-(
I got disconnected due to lightning. My apologies, I am here....
yes, 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
no worries! hope everyone's OK.
No one is outside:) My garage is inside the house, so it's alright.
I wish I had you as my physics teacher, you made everything as easy as even a non-math person such as myself got it!
There 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?`
excellent, 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...
well you know how long it will take to get to Julie, right
ok, i will take a note of pythagorean as regards to velocities as well as distances....
brill! you know ho far apart they are? you know the horizontal velocity? and so?
Horizontal velocity is 9m/s. And the distance between them would be?
didn't you work that out in an earlier part of the question?
use that number
I did the maximum height for [the part in] picture 1, and I did horizontal and vertical displacements for that (and this) part(s).
wait, is it 11cos(44) ?
but that is the velocity of the first through, not the distance. So that is not it.
9(the horiz. velocity of the first through))
let me work it out first, it might jog your memory just wait
OK 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.
oh, so the displacement-for-y formula and displacement-for-x formula are all same, besides that every y is now x instead?
always 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
wait, 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?
we 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.
and 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
While 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).
But, distance I can use that same displacement formula, applying everything I know from picture 1?
no 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
distance = 12.32
meters*
\(12.32 = 9t + \frac{1}{2}(-9.81)t^2\) ?
and olve for time?
solve
http://www.wolframalpha.com/input/?i=2%2811+sin+44%29%2811+cos+44%29+%2F+9.81 correct
OK 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....
Oh, I am setting incorrect equation, because we are working the time based on the information we have from the first question, so u=11
and I said u=9
\(12.32 = 11t + \frac{1}{2}(-9.81)t^2\) But I still get imaginary solutions
with +9.81 I get real solutions:)
just 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?
yes, we had all of this either as given or found that:)
so 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
\(12.327= 9t + \frac{1}{2}(-9.81)t^2\)
there is no gravity in the horizontal direction:p
just constant velocity in horizontal direction
what is my formula for x then?
which formula do I use, I mean
oooHHHHH!!!! think!
\(x = ut + \frac{1}{2}t^2\) ?
the first one you learn!
Distance = Velocity × Time ?
\[x = ut + \frac{1}{2}(0)t^2 = ut\] or t = distance / velocity !!!!
yeah!!
oh, 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 )
s is seconds.
for whatever that quotient is
yes, 1.37s http://www.wolframalpha.com/input/?i=2%2811+sin+44%29%2811+cos+44%29+%2F+%289.81+*+9%29
≈ 1.37
yes:)
Now 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.
one 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?
yes, I undertand that! You are a wonderful person!
so we need an equation of motion with acceleration in it because of gravity \[x = ut + \frac{1}{2}at^2\]
we 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
x 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 \(x = 10.39m/s \times 1.37s + \frac{1}{2}(-9.81m/s^2)(1.37s)^2\)
so 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. \(x = 10.39 \times 1.37 + \frac{1}{2}(-9.81)(1.37)^2\)
and whatever that is equal to....
5.028m
Verif. http://www.wolframalpha.com/input/?i=12.327%3D9t%2B%281%2F2%29%289.81%29t%C2%B2
+1.5m
yes so she throws it 5.03m above the first girls hand. add 1.5m onto that if you want distance above ground
I get 5.028+1.5=6.528
oh! good you got it yourself:p
well, that's the right "way" to do it, any mistakes are calculations.....we now know "how" to do it:p
so the main things I need are: \(\large v^2=u^2+2ax\) \(\large x=ut+\dfrac{1}{2}at^2\)
and of course I know how to find horizontal and vertical velocity components based on a given angle and initial velocity....
let 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.
this is probably the most valuable post in my life. TY so much. No excuse me I got to go:) tnx for formula recap!
yes - horizotal vs vertical there's an awful awful lot of info in there, you have done well! cheers!!
\(v = u + at\) \(v_x~~ =~~u~ \cos(\theta)\) \(v_y ~~=~~ u~\sin(\theta)\) \(d = ut + \frac{1}{2}at^2\) \(v^2 = u^2 + 2ax\) 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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