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|dw:1441289020025:dw|

bsll is supposed to say ball.

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

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

\[v_y = u_y - gt\]
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..

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

no wait, you flipped the sign

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

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

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

oh, -9.81 ... my bad.

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

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.

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.

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 :)

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

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....

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

And that is not correct either -:(

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

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