Ok, can someone show me ....

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Ok, can someone show me ....

Mathematics
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|dw:1441289020025:dw|
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?
bsll is supposed to say ball.

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What's the question ?
How do I find the maximum height that the bAll goes above the ground?
Do you know your equations of motion?
Which one?
I might as well just say, no I don't....
i think we need to know the height of those ppl
at least we need to know the height of the person throwing the ball
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?
f is final, i is initial. just clarifying....
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.
Wait, a is acceleration, but for maximum height, why do we need it? (btw, sorry again and again for deleted the question.)
The key thing to observe is that the vertical component of velocity will be 0 when the ball reaches the maximum height
\[v_y = u_y - gt\] set that equal to 0 and find the time at which that happens
a = g in this case
what is u_y ?
\(a = -g\) because we take upwards as increasing position
\(u_y\) = initial vertical component of velocity
oh. my teacher labelled it \(\LARGE y_{0_{^{_{\huge _x}}}}\)
You can label it what ever, long as it makes sense to you and your teacher
or, idk maybe I am confusing labels. I suck at this:)
thats ridankulous normally we use \(v\) and \(u\) for velocity \(x\) and \(y\) for positions
Yeah it is I that mixed something up (as always)
we never used u, always v_0 xD
Oh yes V_(0y), that!
v = u + at is how i still remember my kinematics eqns from highschool
that's much easier to
it doesn't matter but we all need to agree on same thing..
\(\LARGE V_y=V_{0y}-gt\) \(\LARGE 0=V_{0y}+9.81t\) and what did you say set equal to 0? like this, or am I wrong again?
looks good and do you know why you're setting that equal to 0
no wait, you flipped the sign
\(\LARGE V_y=V_{0y}-gt\) \(\LARGE 0=V_{0y}\color{red}{-}9.81t\) and what did you say set equal to 0? like this, or am I wrong again?
(I just want to do this the way my teacher does it, because I got enough confusion already for 100 seasons in advance....)
oh, you said -g, so I thought... ok will take a note
\(\LARGE 0=V_{0y}+9.81t\)
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)
So we are already having the derivative of the function that we want to maximize. Right?
oh, -9.81 ... my bad.
\(\LARGE 0=V_{0y}-9.81t\)
remember this : \(g\) is acceleration due to gravity, it equals approximately \(9.81\) on surface on earth, it is always positive.
wait, so no minus; plus?
\(a\) is acceleration, in our case \(a = -g = -9.81\)
\(\Large 0=V_{0y}\color{red}{\bf-}9.81t\)
so g is pos, -g is neg. at least I know that. I am a pes(t) - imist.
\(a\) is negative because \(g\) is pointing downward but we are taking "upwards" as increasing position
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)
right, one thing at a time first find the time at which vertical component of velocity becomes 0
t=0?
at t=0 ... V_(0y) is 0.
really ? at \(t=0\) the ball surely has nonzero vertical component of velocity : |dw:1441290728442:dw|
Wait, what is my equation for this? (to answer your question about the t-value at which the vertical velocity is equivalent to zero.)
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 \[V_y=11\sin(44)-9.81t \]
oh, can I try?
let me finish typing first
sure go ahead
il let u work it but im not finished yet and i think u need to get some clarifications :)
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
once you have the "time" at which ball reaches the maximum height, you can use that to find the required maximum height
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
that is all i wanted to say go ahead, set \(V_y\) equal to \(0\) and solve \(t\)
\(\LARGE V_y=11\sin(44)-9.81t \) \(\LARGE 0=11\sin(44)-9.81t \) \(\LARGE 11\sin(44)=9.81t \) \(\LARGE t=\frac{9.81}{11\sin(44)} \)
hope at least that I have managed:)
Hmm
oh no
well that to the power of -1
It should be \[t = \frac{ 11\sin(44) }{ 9.81 }\]
\(\LARGE t=\frac{11\sin(44)}{9.81} \) I am even dumber than what I thought
yes, I am dividing by 9.81 don't know what's up with my Brain
\(\LARGE t=\frac{11\sin(44)}{9.81} \approx 0.78 \)
You could actually go ahead and use \[y = v_{0y} t\] now
That will give you your maximum height
\(\LARGE y=V_{0y}\cdot t =11\sin(44)\cdot 0.78 =5.96\)
calculations did wolfram not me, so that is supposed to be correct if I plugged correctly.
That is the max height....
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 \(\LARGE y=V_{0t}\times t \) \(\LARGE y=11\sin(44)\times 0.78=5.96 \)
t=0?
sorry, what do you mean?
Why did you write max height is at t = 0
oh, occurs at t=0.78
Don't round until the final answer
So use the full time rather than 0.78
\(\LARGE y=\sin^2(44)/9.81\) ?
I forgot 11
0.78 was really 11sin(44)/9.81 So y is y=11sin(44)×11sin(44)÷9.81
well, if that is right then it is still 5.95
\(\Large y = v_{0y} t - \frac{ 1 }{ 2 } g t^2\) \(\Large y = 11\sin(44)\left(\frac{11\sin(44) }{9.81}\right) - \frac{ 1 }{ 2 } (9.81) \left(\frac{11\sin(44) }{9.81}\right) ^2\) 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
And that is not correct either -:(
Hold on, let just use \[h = \frac{ (v_{0y} \sin \theta)^2 }{ (2g) }\] we also have to remember the 1.5m forgot about that
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
I have to go, I will come back about 2 h later-;(
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
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!
below equation for vertical position perfectly fine \(\Large y = \color{red}{y_0}+v_{0y} t - \frac{ 1 }{ 2 } g t^2\) \(\Large y = \color{red}{y_0}+11\sin(44)\left(\frac{11\sin(44) }{9.81}\right) - \frac{ 1 }{ 2 } (9.81) \left(\frac{11\sin(44) }{9.81}\right) ^2\)
When you put your final answer, make sure it's 2 sig digs!
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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