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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?

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  1. Venomblast
    • one year ago
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    I got the formula x=x(initial)=v(inital x)*t+1/2at^2

  2. anonymous
    • one year ago
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    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

  3. anonymous
    • one year ago
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    Your formula is incorrect because it is a 2 dimensional motion (x and y component changes) You can ignore the height of the volcano

  4. anonymous
    • one year ago
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    But to find R convert the horizontal distance of km to meters

  5. Venomblast
    • one year ago
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    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 \[V _{0x}= v _{0}*\cos(45)\] \[V _{0y}= v _{0}*\sin(45)\]

  6. Venomblast
    • one year ago
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    besides I don't know what's R and the initial velocity

  7. anonymous
    • one year ago
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    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?

  8. Venomblast
    • one year ago
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    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?

  9. anonymous
    • one year ago
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    \[v _{xi}=v _{i}\cos \theta \] \[v _{yi}=v _{i}\sin \theta \]

  10. Venomblast
    • one year ago
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    I did that. But what V initial?"

  11. anonymous
    • one year ago
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    Thats what I said. You mentioned the formula I wrote and I told you that it wasnt needed.

  12. Venomblast
    • one year ago
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    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/

  13. Venomblast
    • one year ago
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    so the v( y initial) is irrelevant?

  14. anonymous
    • one year ago
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    You sure? Check your notes or online lessons?

  15. Venomblast
    • one year ago
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    yes. for a thrown object, a=g v=gt+v x=1/2 t^2 + v(initial)

  16. anonymous
    • one year ago
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    Its a projectile motion. |dw:1443681041543:dw| It is 2D dimensional

  17. Venomblast
    • one year ago
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    and for the y component, \[y _{0}+V _{_{0y}}+\frac{ 1}{ 2 } a _{y}t ^{2}\]

  18. Venomblast
    • one year ago
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    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?

  19. anonymous
    • one year ago
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    It doesnt work that way. The y component only works on 1d This is 2D |dw:1443681204247:dw|

  20. Venomblast
    • one year ago
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    I dont ge tit

  21. anonymous
    • one year ago
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    Follow my instruction by using the formula I mentioned in the beginning and you will.

  22. Venomblast
    • one year ago
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    but the initial velocity is still unknown. I want to learn it by solving velocity with the projectile motion.

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

  24. anonymous
    • one year ago
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    \[t=\frac{ v _{i}\sin \theta }{ g }\] since \[v _{yf}=v _{yi}-g*t\] \[0=v _{i}\sin \theta-g*t\]

  25. anonymous
    • one year ago
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    So \[h=\frac{ v _{i}^2\sin^2\theta }{ 2g }\]

  26. Venomblast
    • one year ago
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    isn't it g^2?

  27. anonymous
    • one year ago
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    But that equation is useless. We dont know the height from the volcano to the highest point.

  28. anonymous
    • one year ago
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    So instead we use this one I mentioned earlier.

  29. Venomblast
    • one year ago
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    it 3.3km

  30. anonymous
    • one year ago
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    \[x _{f}=x _{i}+v _{x i}t \rightarrow R=v _{x i}t \rightarrow (v _{i}\cos \theta)2t\]

  31. anonymous
    • one year ago
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    \[(v _{i}\cos \theta)\frac{ 2v _{i}\sin \theta }{ g }=\frac{ 2v _{i}^2\sin \theta \cos \theta }{ g }\]

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

  33. Venomblast
    • one year ago
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    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

  34. anonymous
    • one year ago
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    h is the max height it reaches at its starting point. We dont know that. |dw:1443682444669:dw|

  35. anonymous
    • one year ago
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    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.

  36. Venomblast
    • one year ago
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    it erupt. there must be some speed for the vertical but how is there an innital for the horizontal component?

  37. anonymous
    • one year ago
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    Its a projectile motion. It goes both ways.

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

  39. Venomblast
    • one year ago
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    ok.. So how do i get velocity?

  40. Venomblast
    • one year ago
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    @Shalante what v then?

  41. anonymous
    • one year ago
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    Look at my first posts.

  42. Venomblast
    • one year ago
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    thanks. It is still confusing. I just trying to make sense in my perspective

  43. anonymous
    • one year ago
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    R has to be meters. The horizontal distance of 9.4 km is kilometers. Did you convert it?

  44. anonymous
    • one year ago
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    What are you confused at?

  45. IrishBoy123
    • one year ago
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    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: \(-3300 = 9400 + \frac{1}{2} (-9.8) t^2\)

  46. Venomblast
    • one year ago
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    iRISHBOY How did you get v/sqrt(2)? I got the solution but i don't get what my professor did

  47. Venomblast
    • one year ago
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    How do you solve this type of problem. I am taking quantitative physic

  48. anonymous
    • one year ago
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    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 \[v _{x i}t=9400m\] \[v _{yi}t \neq v _{x i}t =9400m\]

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

  50. IrishBoy123
    • one year ago
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    @Venomblast \(\cos 45 = \sin 45 = \dfrac{1}{\sqrt{2}}\)

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

  52. Venomblast
    • one year ago
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    What do I do now?

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