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|dw:1441307711913:dw||dw:1441308026115:dw|

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

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?

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
\[\large 0^2=u^2+2(-9.81)(5.5)\]

yes, if I plugged in +9.81 I would have got imaginary velocity.

u=10.39 m/s (this is initial velocy)

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

I got disconnected due to lightning. My apologies, I am here....

no worries! hope everyone's OK.

No one is outside:) My garage is inside the house, so it's alright.

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

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

But, distance I can use that same displacement formula, applying everything I know from picture 1?

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

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:)

yes, we had all of this either as given or found that:)

\(12.327= 9t + \frac{1}{2}(-9.81)t^2\)

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!!

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:)

yes, I undertand that! You are a wonderful person!

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

I get
5.028+1.5=6.528

oh! good you got it yourself:p

so the main things I need are:
\(\large v^2=u^2+2ax\)
\(\large x=ut+\dfrac{1}{2}at^2\)