A volcanic block is ejected at an angle of 45 degrees from Mount Fuji during a volcanic eruption. It lands at the foot of the volcano at an horizontal distance of 9.4 km. The height of Mount Fuji is 3.3 km. What is the block's initial speed?
Do I use Pythagorean theorem to find the hypotenuse?
Is my distance known?

- Venomblast

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- Venomblast

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

- anonymous

Nope you would use this formula
\[R=\frac{ v _{i}^2\sin2\theta }{ g }\]
where R is the horizontal distance it traveled
vi is intial velocity
theta is the angle
g is the acceleration of gravity

- anonymous

Your formula is incorrect because it is a 2 dimensional motion (x and y component changes)
You can ignore the height of the volcano

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- Venomblast

The think is I never learned that formula/ This is a calc base physic. So I need to to get the velocity by figuring out what I have. I drew a triangle with the angle
I get \[V _{0x}= v _{0}*\cos(45)\]
\[V _{0y}= v _{0}*\sin(45)\]

- Venomblast

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

- anonymous

You would need initial velocity to use that formula of yours.
It would be the x and y component of the initial velocity.
Where were you given "that" in the question?

- Venomblast

there is not a initial velocity. This is suppose to make you find it. So I was thinking. when t(time) is 1/2 does that equal 0 for the y component since it is tangible half way making no velocity when time reaches 1 half of it's destination?

- anonymous

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

- Venomblast

I did that. But what V initial?"

- anonymous

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

- Venomblast

But I dont think my professor wants to see that since I have not learned it yet. He want's me to use what I have which is the distance formula and velocity/

- Venomblast

so the v( y initial) is irrelevant?

- anonymous

You sure?
Check your notes or online lessons?

- Venomblast

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

- anonymous

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

- Venomblast

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

- Venomblast

Yes I know. so I need to take account of gravity. SO if it takes x(time) to reach 9.4 km. In half time it reached half it's distance correct?

- anonymous

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

- Venomblast

I dont ge tit

- anonymous

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

- Venomblast

but the initial velocity is still unknown. I want to learn it by solving velocity with the projectile motion.

- anonymous

Look.
\[h=y _{f}\]
\[h=y _{i}+v _{yi}t-\frac{ 1 }{ 2 }g*t^2\]
\[h=v _{i}\sin \theta \times \frac{ v _{i}\sin \theta }{ g }-\frac{ 1 }{ 2 }g*(\frac{ v _{i}\sin \theta }{ g })^2\]

- anonymous

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

- anonymous

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

- Venomblast

isn't it g^2?

- anonymous

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

- anonymous

So instead we use this one I mentioned earlier.

- Venomblast

it 3.3km

- anonymous

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

- anonymous

\[(v _{i}\cos \theta)\frac{ 2v _{i}\sin \theta }{ g }=\frac{ 2v _{i}^2\sin \theta \cos \theta }{ g }\]

- anonymous

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

- Venomblast

x(initial) is zero since it is not given. There was some acceleration so all I am left with is v*a*t^2 +v(initial)*t

- anonymous

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

- anonymous

45 angles thrown off from a cliff is the same distance as 45 angles thrown from the ground.
So height of the volcano is irrelevant.

- Venomblast

it erupt. there must be some speed for the vertical but how is there an innital for the horizontal component?

- anonymous

Its a projectile motion.
It goes both ways.

- anonymous

All I can say is use my formula with the R.
If you're not going to listen to me, why ask the question?

- Venomblast

ok.. So how do i get velocity?

- Venomblast

@Shalante what v then?

- anonymous

Look at my first posts.

- Venomblast

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

- anonymous

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

- anonymous

What are you confused at?

- IrishBoy123

the height of the volcano is certainly not irrelevant in this question
you have correctly identified that, if the initial velocity of the rock is v, then it's horizontal and vertical components are both \(\dfrac{v}{\sqrt{2}}\)
for total flight time t, in the horizontal direction you can say that \(\frac{v}{\sqrt{2}} t= 9400\) as velocity is assumed to be constant [no air resistance etc]
in the vertical direction, using your equation \(x = v_{i}t + \frac{1}{2}at^2\) we can say that \(-3300 = \frac{v}{\sqrt{2}} t + \frac{1}{2} (-9.8) t^2\) [positive is up, ie note signs]
that simplifies nicely:
\(-3300 = 9400 + \frac{1}{2} (-9.8) t^2\)

- Venomblast

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

- Venomblast

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

- anonymous

No.
The x and y components are separated and they are 2 different equation
\[y _{f}=y _{i}+v _{y i}t+\frac{ 1 }{ 2 }a _{y}t^2\]
\[x _{f}=x _{i}+v _{x i}t+\frac{ 1 }{ 2 }a _{x} t^2\]
Sure, I agree that \[(v _{x i}\cos45)t\] equals\[\frac{ v }{ \sqrt{2} }t=9400 \]
but
If the acceleration is gravity (9.8m/s^2) then this formula have to be in y component, including the height (yf): \[y _{f}=y _{i}+v _{yi}t+\frac{ 1 }{ 2 }a _{y}t^2\]
and \[v _{x i}t=9400m\]
\[v _{yi}t \neq v _{x i}t =9400m\]

- anonymous

\[x _{f}= v _{x i}t\] if velocity of horizontal component is constant.
\[y _{f}=v _{y i}t\] The acceleration always changes in the y direction. (gravity)
So \[x _{f}\neq y _{f}\]

- IrishBoy123

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

- Venomblast

\[\tan(45)=\frac{ v _{y} }{ v _{x} } => 1=\frac{ v _{y} }{ v _{x} } \]
Thus they're both equal. Ok what do I do now? I will upload my work

- Venomblast

What do I do now?

##### 1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.