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Venomblast
 one year ago
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
 one year ago
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?

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Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0I got the formula x=x(initial)=v(inital x)*t+1/2at^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope 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
 one year ago
Best ResponseYou've already chosen the best response.0Your formula is incorrect because it is a 2 dimensional motion (x and y component changes) You can ignore the height of the volcano

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But to find R convert the horizontal distance of km to meters

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0besides I don't know what's R and the initial velocity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.0there 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
 one year ago
Best ResponseYou've already chosen the best response.0\[v _{xi}=v _{i}\cos \theta \] \[v _{yi}=v _{i}\sin \theta \]

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0I did that. But what V initial?"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thats what I said. You mentioned the formula I wrote and I told you that it wasnt needed.

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0But 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
 one year ago
Best ResponseYou've already chosen the best response.0so the v( y initial) is irrelevant?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You sure? Check your notes or online lessons?

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0yes. for a thrown object, a=g v=gt+v x=1/2 t^2 + v(initial)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its a projectile motion. dw:1443681041543:dw It is 2D dimensional

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0and for the y component, \[y _{0}+V _{_{0y}}+\frac{ 1}{ 2 } a _{y}t ^{2}\]

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0Yes 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
 one year ago
Best ResponseYou've already chosen the best response.0It doesnt work that way. The y component only works on 1d This is 2D dw:1443681204247:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Follow my instruction by using the formula I mentioned in the beginning and you will.

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0but the initial velocity is still unknown. I want to learn it by solving velocity with the projectile motion.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Look. \[h=y _{f}\] \[h=y _{i}+v _{yi}t\frac{ 1 }{ 2 }g*t^2\] \[h=v _{i}\sin \theta \times \frac{ v _{i}\sin \theta }{ g }\frac{ 1 }{ 2 }g*(\frac{ v _{i}\sin \theta }{ g })^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[t=\frac{ v _{i}\sin \theta }{ g }\] since \[v _{yf}=v _{yi}g*t\] \[0=v _{i}\sin \thetag*t\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So \[h=\frac{ v _{i}^2\sin^2\theta }{ 2g }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But that equation is useless. We dont know the height from the volcano to the highest point.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So instead we use this one I mentioned earlier.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x _{f}=x _{i}+v _{x i}t \rightarrow R=v _{x i}t \rightarrow (v _{i}\cos \theta)2t\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(v _{i}\cos \theta)\frac{ 2v _{i}\sin \theta }{ g }=\frac{ 2v _{i}^2\sin \theta \cos \theta }{ g }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[R=\frac{ v _{i}^2\sin2\theta }{ g }\] R should be in meters, so convert it first.

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0x(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
 one year ago
Best ResponseYou've already chosen the best response.0h is the max height it reaches at its starting point. We dont know that. dw:1443682444669:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.045 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
 one year ago
Best ResponseYou've already chosen the best response.0it erupt. there must be some speed for the vertical but how is there an innital for the horizontal component?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Its a projectile motion. It goes both ways.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0All I can say is use my formula with the R. If you're not going to listen to me, why ask the question?

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0ok.. So how do i get velocity?

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0@Shalante what v then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Look at my first posts.

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0thanks. It is still confusing. I just trying to make sense in my perspective

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0R has to be meters. The horizontal distance of 9.4 km is kilometers. Did you convert it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What are you confused at?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0iRISHBOY How did you get v/sqrt(2)? I got the solution but i don't get what my professor did

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0How do you solve this type of problem. I am taking quantitative physic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No. 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
 one year ago
Best ResponseYou've already chosen the best response.0\[x _{f}= v _{x i}t\] 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
 one year ago
Best ResponseYou've already chosen the best response.0@Venomblast \(\cos 45 = \sin 45 = \dfrac{1}{\sqrt{2}}\)

Venomblast
 one year ago
Best ResponseYou've already chosen the best response.0\[\tan(45)=\frac{ v _{y} }{ v _{x} } => 1=\frac{ v _{y} }{ v _{x} } \] Thus they're both equal. Ok what do I do now? I will upload my work
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