idku
  • idku
Ok, can someone show me ....
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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idku
  • idku
|dw:1441289020025:dw|
idku
  • idku
I found: Horizontal component of the velocity: \(\LARGE V_x=11\cos(44)\approx 7.9\) Vertical component of the velocity: \(\LARGE V_x=11\sin(44)\approx 7.6\) ---------------------------------------------------- How do I find the maximum height that the bsll goes above the ground?
idku
  • idku
bsll is supposed to say ball.

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ganeshie8
  • ganeshie8
What's the question ?
idku
  • idku
How do I find the maximum height that the bAll goes above the ground?
anonymous
  • anonymous
Do you know your equations of motion?
idku
  • idku
Which one?
idku
  • idku
I might as well just say, no I don't....
ganeshie8
  • ganeshie8
i think we need to know the height of those ppl
ganeshie8
  • ganeshie8
at least we need to know the height of the person throwing the ball
anonymous
  • anonymous
This is projectile motion correct, so it's better if you label your vector quantities more precisely, so we're dealing with the y - direction here\[\vec a = \frac{ v_{yf}-v_{yi} }{ t }\] where \[v_{yf}=0\] is there more information to this, all you gave is a diagram?
idku
  • idku
f is final, i is initial. just clarifying....
idku
  • idku
I deleted the question I am so bump ---------------------------- Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 11 m/s at an angle 44 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
idku
  • idku
Wait, a is acceleration, but for maximum height, why do we need it? (btw, sorry again and again for deleted the question.)
ganeshie8
  • ganeshie8
The key thing to observe is that the vertical component of velocity will be 0 when the ball reaches the maximum height
ganeshie8
  • ganeshie8
\[v_y = u_y - gt\] set that equal to 0 and find the time at which that happens
anonymous
  • anonymous
a = g in this case
idku
  • idku
what is u_y ?
ganeshie8
  • ganeshie8
\(a = -g\) because we take upwards as increasing position
ganeshie8
  • ganeshie8
\(u_y\) = initial vertical component of velocity
idku
  • idku
oh. my teacher labelled it \(\LARGE y_{0_{^{_{\huge _x}}}}\)
anonymous
  • anonymous
You can label it what ever, long as it makes sense to you and your teacher
idku
  • idku
or, idk maybe I am confusing labels. I suck at this:)
ganeshie8
  • ganeshie8
thats ridankulous normally we use \(v\) and \(u\) for velocity \(x\) and \(y\) for positions
idku
  • idku
Yeah it is I that mixed something up (as always)
anonymous
  • anonymous
we never used u, always v_0 xD
idku
  • idku
Oh yes V_(0y), that!
ganeshie8
  • ganeshie8
v = u + at is how i still remember my kinematics eqns from highschool
anonymous
  • anonymous
that's much easier to
ganeshie8
  • ganeshie8
it doesn't matter but we all need to agree on same thing..
idku
  • idku
\(\LARGE V_y=V_{0y}-gt\) \(\LARGE 0=V_{0y}+9.81t\) and what did you say set equal to 0? like this, or am I wrong again?
ganeshie8
  • ganeshie8
looks good and do you know why you're setting that equal to 0
ganeshie8
  • ganeshie8
no wait, you flipped the sign
ganeshie8
  • ganeshie8
\(\LARGE V_y=V_{0y}-gt\) \(\LARGE 0=V_{0y}\color{red}{-}9.81t\) and what did you say set equal to 0? like this, or am I wrong again?
idku
  • idku
(I just want to do this the way my teacher does it, because I got enough confusion already for 100 seasons in advance....)
idku
  • idku
oh, you said -g, so I thought... ok will take a note
idku
  • idku
\(\LARGE 0=V_{0y}+9.81t\)
idku
  • idku
And I am setting it to 0, because V_y is the derivative of D_y (where D is displacement - in ase of y, the height)
idku
  • idku
So we are already having the derivative of the function that we want to maximize. Right?
idku
  • idku
oh, -9.81 ... my bad.
idku
  • idku
\(\LARGE 0=V_{0y}-9.81t\)
ganeshie8
  • ganeshie8
remember this : \(g\) is acceleration due to gravity, it equals approximately \(9.81\) on surface on earth, it is always positive.
idku
  • idku
wait, so no minus; plus?
ganeshie8
  • ganeshie8
\(a\) is acceleration, in our case \(a = -g = -9.81\)
idku
  • idku
\(\Large 0=V_{0y}\color{red}{\bf-}9.81t\)
idku
  • idku
so g is pos, -g is neg. at least I know that. I am a pes(t) - imist.
ganeshie8
  • ganeshie8
\(a\) is negative because \(g\) is pointing downward but we are taking "upwards" as increasing position
idku
  • idku
Ok, and then I just find cthe maximum of our unknown displacement of y, based on the given velocity component (which is the derivative of the displacement)
ganeshie8
  • ganeshie8
right, one thing at a time first find the time at which vertical component of velocity becomes 0
idku
  • idku
t=0?
idku
  • idku
at t=0 ... V_(0y) is 0.
ganeshie8
  • ganeshie8
really ? at \(t=0\) the ball surely has nonzero vertical component of velocity : |dw:1441290728442:dw|
idku
  • idku
Wait, what is my equation for this? (to answer your question about the t-value at which the vertical velocity is equivalent to zero.)
ganeshie8
  • ganeshie8
Easy, as you said, vertical component of velocity of ball is modeled by the function : \[V_y=V_{0y}-9.81t \] you're "given" \(V_{0y} = 11\sin(44)\), therefore the function becomes \[V_y=11\sin(44)-9.81t \]
idku
  • idku
oh, can I try?
ganeshie8
  • ganeshie8
let me finish typing first
idku
  • idku
sure go ahead
ganeshie8
  • ganeshie8
il let u work it but im not finished yet and i think u need to get some clarifications :)
ganeshie8
  • ganeshie8
you know that \(V_y\) is \(0\) when the ball reaches maximum height so use that fact to find the "time" at which the ball reaches maximum height
ganeshie8
  • ganeshie8
once you have the "time" at which ball reaches the maximum height, you can use that to find the required maximum height
anonymous
  • anonymous
This is a nice way to work with vertical trajectory you could also still do it the way I was showing above, the main thing is knowing the final velocity of the y component = 0 at maximum height
ganeshie8
  • ganeshie8
that is all i wanted to say go ahead, set \(V_y\) equal to \(0\) and solve \(t\)
idku
  • idku
\(\LARGE V_y=11\sin(44)-9.81t \) \(\LARGE 0=11\sin(44)-9.81t \) \(\LARGE 11\sin(44)=9.81t \) \(\LARGE t=\frac{9.81}{11\sin(44)} \)
idku
  • idku
hope at least that I have managed:)
anonymous
  • anonymous
Hmm
idku
  • idku
oh no
idku
  • idku
well that to the power of -1
anonymous
  • anonymous
It should be \[t = \frac{ 11\sin(44) }{ 9.81 }\]
idku
  • idku
\(\LARGE t=\frac{11\sin(44)}{9.81} \) I am even dumber than what I thought
idku
  • idku
yes, I am dividing by 9.81 don't know what's up with my Brain
idku
  • idku
\(\LARGE t=\frac{11\sin(44)}{9.81} \approx 0.78 \)
anonymous
  • anonymous
You could actually go ahead and use \[y = v_{0y} t\] now
anonymous
  • anonymous
That will give you your maximum height
idku
  • idku
\(\LARGE y=V_{0y}\cdot t =11\sin(44)\cdot 0.78 =5.96\)
idku
  • idku
calculations did wolfram not me, so that is supposed to be correct if I plugged correctly.
idku
  • idku
That is the max height....
idku
  • idku
Let me make an attempt to recap what I have done for maximum height ((My suffering is not equivalent of any sin (of \(\theta\)), I hate this!)) ----------------------------------------------------- RECAP, MAX HEIGHT \(\LARGE V_y=V_{0y}-gt \) \(\LARGE 0=11\sin(44)-9.81t \) \(\LARGE t =0.78\) so max height occrs at t=0, and now we plug that into vertical displacement \(\LARGE y=V_{0t}\times t \) \(\LARGE y=11\sin(44)\times 0.78=5.96 \)
anonymous
  • anonymous
t=0?
idku
  • idku
sorry, what do you mean?
anonymous
  • anonymous
Why did you write max height is at t = 0
idku
  • idku
oh, occurs at t=0.78
anonymous
  • anonymous
Don't round until the final answer
anonymous
  • anonymous
So use the full time rather than 0.78
idku
  • idku
\(\LARGE y=\sin^2(44)/9.81\) ?
idku
  • idku
I forgot 11
idku
  • idku
0.78 was really 11sin(44)/9.81 So y is y=11sin(44)×11sin(44)÷9.81
idku
  • idku
well, if that is right then it is still 5.95
idku
  • idku
\(\Large y = v_{0y} t - \frac{ 1 }{ 2 } g t^2\) \(\Large y = 11\sin(44)\left(\frac{11\sin(44) }{9.81}\right) - \frac{ 1 }{ 2 } (9.81) \left(\frac{11\sin(44) }{9.81}\right) ^2\) http://www.wolframalpha.com/input/?i=11sin%2844%29%C3%97%2811sin%2844%29%2F9.81%29-%281%2F2%29%289.81%29%2811sin%2844%29%2F9.81%29%C2%B2 I get 2.97597 ----> 2.98
idku
  • idku
And that is not correct either -:(
anonymous
  • anonymous
Hold on, let just use \[h = \frac{ (v_{0y} \sin \theta)^2 }{ (2g) }\] we also have to remember the 1.5m forgot about that
idku
  • idku
I don't think that there is another explanation for an incorrect output from the formula, other than that I failed to find all the values correctly. I don't think formulas are faulty. (I start to despair from thinking that I can think ) HIHPHh9hduhoduhdue89wv9vi
idku
  • idku
I have to go, I will come back about 2 h later-;(
anonymous
  • anonymous
Yes, this should work, the problem was we did not take account of the question, you must read it carefully, and not make assumptions. Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 11 m/s at an angle 44 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground. \[h = \frac{ (v_{0y} \sin \theta)^2 }{ (2g) } = \frac{ [11\sin(44)]^2 }{ 2(9.81)} = 2.975...\] but we must add the 1.5m as well to this, so we get 2.975...+1.5 = 4.4759m = 4.5m
anonymous
  • anonymous
I think you should use this though as we found time for this precisely \[h = \frac{ 1 }{ 2 }g t^2 = \frac{ 1 }{ 2 }(9.81)(0.78)^2 = 2.965....+1.5 = 4.47 = 4.5m\] so it works!
ganeshie8
  • ganeshie8
below equation for vertical position perfectly fine \(\Large y = \color{red}{y_0}+v_{0y} t - \frac{ 1 }{ 2 } g t^2\) \(\Large y = \color{red}{y_0}+11\sin(44)\left(\frac{11\sin(44) }{9.81}\right) - \frac{ 1 }{ 2 } (9.81) \left(\frac{11\sin(44) }{9.81}\right) ^2\)
anonymous
  • anonymous
When you put your final answer, make sure it's 2 sig digs!
idku
  • idku
Oh, if it was something on a same level of thinking then I would have probably figured. Yes the 1.5 meters of the ground, because my origin is (0, 1.5) Iambatman tnx! I am checking it and it is.....

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