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I got the formula x=x(initial)=v(inital x)*t+1/2at^2

But to find R convert the horizontal distance of km to meters

besides I don't know what's R and the initial velocity

\[v _{xi}=v _{i}\cos \theta \]
\[v _{yi}=v _{i}\sin \theta \]

I did that. But what V initial?"

Thats what I said.
You mentioned the formula I wrote and I told you that it wasnt needed.

so the v( y initial) is irrelevant?

You sure?
Check your notes or online lessons?

yes.
for a thrown object,
a=g
v=gt+v
x=1/2 t^2 + v(initial)

Its a projectile motion.
|dw:1443681041543:dw|
It is 2D dimensional

and for the y component,
\[y _{0}+V _{_{0y}}+\frac{ 1}{ 2 } a _{y}t ^{2}\]

It doesnt work that way.
The y component only works on 1d
This is 2D
|dw:1443681204247:dw|

I dont ge tit

Follow my instruction by using the formula I mentioned in the beginning and you will.

\[t=\frac{ v _{i}\sin \theta }{ g }\]
since \[v _{yf}=v _{yi}-g*t\]
\[0=v _{i}\sin \theta-g*t\]

So \[h=\frac{ v _{i}^2\sin^2\theta }{ 2g }\]

isn't it g^2?

But that equation is useless.
We dont know the height from the volcano to the highest point.

So instead we use this one I mentioned earlier.

it 3.3km

\[x _{f}=x _{i}+v _{x i}t \rightarrow R=v _{x i}t \rightarrow (v _{i}\cos \theta)2t\]

\[R=\frac{ v _{i}^2\sin2\theta }{ g }\]
R should be in meters, so convert it first.

h is the max height it reaches at its starting point.
We dont know that.
|dw:1443682444669:dw|

Its a projectile motion.
It goes both ways.

ok.. So how do i get velocity?

@Shalante what v then?

Look at my first posts.

thanks. It is still confusing. I just trying to make sense in my perspective

R has to be meters.
The horizontal distance of 9.4 km is kilometers.
Did you convert it?

What are you confused at?

iRISHBOY How did you get v/sqrt(2)? I got the solution but i don't get what my professor did

How do you solve this type of problem. I am taking quantitative physic

@Venomblast
\(\cos 45 = \sin 45 = \dfrac{1}{\sqrt{2}}\)

What do I do now?