Ok, can someone show me ....

- idku

Ok, can someone show me ....

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- idku

|dw:1441289020025:dw|

- 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

bsll is supposed to say ball.

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## More answers

- ganeshie8

What's the question ?

- idku

How do I find the maximum height that the bAll goes above the ground?

- anonymous

Do you know your equations of motion?

- idku

Which one?

- idku

I might as well just say, no I don't....

- ganeshie8

i think we need to know the height of those ppl

- ganeshie8

at least we need to know the height of the person throwing the ball

- 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

f is final, i is initial.
just clarifying....

- 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

Wait, a is acceleration, but for maximum height, why do we need it?
(btw, sorry again and again for deleted the question.)

- ganeshie8

The key thing to observe is that the vertical component of velocity will be 0 when the ball reaches the maximum height

- ganeshie8

\[v_y = u_y - gt\]
set that equal to 0 and find the time at which that happens

- anonymous

a = g in this case

- idku

what is u_y ?

- ganeshie8

\(a = -g\)
because we take upwards as increasing position

- ganeshie8

\(u_y\) = initial vertical component of velocity

- idku

oh. my teacher labelled it \(\LARGE y_{0_{^{_{\huge _x}}}}\)

- anonymous

You can label it what ever, long as it makes sense to you and your teacher

- idku

or, idk maybe I am confusing labels. I suck at this:)

- ganeshie8

thats ridankulous
normally we use \(v\) and \(u\) for velocity
\(x\) and \(y\) for positions

- idku

Yeah it is I that mixed something up (as always)

- anonymous

we never used u, always v_0 xD

- idku

Oh yes V_(0y), that!

- ganeshie8

v = u + at
is how i still remember my kinematics eqns from highschool

- anonymous

that's much easier to

- ganeshie8

it doesn't matter
but we all need to agree on same thing..

- 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

looks good
and do you know why you're setting that equal to 0

- ganeshie8

no wait, you flipped the sign

- 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

(I just want to do this the way my teacher does it, because I got enough confusion already for 100 seasons in advance....)

- idku

oh, you said -g, so I thought... ok will take a note

- idku

\(\LARGE 0=V_{0y}+9.81t\)

- 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

So we are already having the derivative of the function that we want to maximize.
Right?

- idku

oh, -9.81 ... my bad.

- idku

\(\LARGE 0=V_{0y}-9.81t\)

- ganeshie8

remember this :
\(g\) is acceleration due to gravity, it equals approximately \(9.81\) on surface on earth, it is always positive.

- idku

wait, so no minus; plus?

- ganeshie8

\(a\) is acceleration, in our case \(a = -g = -9.81\)

- idku

\(\Large 0=V_{0y}\color{red}{\bf-}9.81t\)

- idku

so g is pos, -g is neg.
at least I know that. I am a pes(t) - imist.

- ganeshie8

\(a\) is negative because \(g\) is pointing downward
but we are taking "upwards" as increasing position

- 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

right, one thing at a time
first find the time at which vertical component of velocity becomes 0

- idku

t=0?

- idku

at t=0 ... V_(0y) is 0.

- ganeshie8

really ?
at \(t=0\) the ball surely has nonzero vertical component of velocity :
|dw:1441290728442:dw|

- 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

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

oh, can I try?

- ganeshie8

let me finish typing first

- idku

sure go ahead

- ganeshie8

il let u work it
but im not finished yet and i think u need to get some clarifications :)

- 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

once you have the "time" at which ball reaches the maximum height,
you can use that to find the required maximum height

- 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

that is all i wanted to say
go ahead, set \(V_y\) equal to \(0\) and solve \(t\)

- 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

hope at least that I have managed:)

- anonymous

Hmm

- idku

oh no

- idku

well that to the power of -1

- anonymous

It should be \[t = \frac{ 11\sin(44) }{ 9.81 }\]

- idku

\(\LARGE t=\frac{11\sin(44)}{9.81} \)
I am even dumber than what I thought

- idku

yes, I am dividing by 9.81
don't know what's up with my Brain

- idku

\(\LARGE t=\frac{11\sin(44)}{9.81} \approx 0.78 \)

- anonymous

You could actually go ahead and use \[y = v_{0y} t\] now

- anonymous

That will give you your maximum height

- idku

\(\LARGE y=V_{0y}\cdot t =11\sin(44)\cdot 0.78 =5.96\)

- idku

calculations did wolfram not me, so that is supposed to be correct if I plugged correctly.

- idku

That is the max height....

- 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

t=0?

- idku

sorry, what do you mean?

- anonymous

Why did you write max height is at t = 0

- idku

oh, occurs at t=0.78

- anonymous

Don't round until the final answer

- anonymous

So use the full time rather than 0.78

- idku

\(\LARGE y=\sin^2(44)/9.81\) ?

- idku

I forgot 11

- idku

0.78 was really 11sin(44)/9.81
So y is
y=11sin(44)×11sin(44)÷9.81

- idku

well, if that is right then it is still 5.95

- 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

And that is not correct either -:(

- 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

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

I have to go, I will come back about 2 h later-;(

- 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

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

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

When you put your final answer, make sure it's 2 sig digs!

- 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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