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zaphod Group Title

A particle is projected with speed v at an angle alpha to the horizontal. Find the speed of the particle when it is at height h

  • 2 years ago
  • 2 years ago

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  1. zaphod Group Title
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    @UnkleRhaukus @ash2326

    • 2 years ago
  2. ash2326 Group Title
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    @zaphod Particle's velocity has two components : 1) Horizontal Velocity= \(v\cos \alpha\) {remains unchanged throughout the length of the motion obviously we are ignoring the air friction } 2) Vertical Velocity=\(v \sin \alpha\) initially { decelerated by gravity} |dw:1349870216959:dw| Do you understand this part?

    • 2 years ago
  3. zaphod Group Title
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    yes now how do i get the height into this can u show the working

    • 2 years ago
  4. ash2326 Group Title
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    Only \( v \sin \alpha\) is used for vertical motion. Using \[v^2-u^2=2as\] Taking upward direction to be positive We need to find vertical velocity when the particle reaches a height h \[v^2-u^2=2as\] s= h a=-g \[u=v\sin \alpha \] \[v^2-(u\sin \alpha)^2=2\times (-g)\times h\] now find v from this This will be the vertical velocity at height h |dw:1349870813000:dw|

    • 2 years ago
  5. zaphod Group Title
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    speed will be the resultant of v cos alpha and wt u found out?

    • 2 years ago
  6. ash2326 Group Title
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    Note that this v is different from the v given in the question. Let it be x \[x|dw:1349871003737:dw|^2-(u\sin \alpha)^2=2\times (-g)\times h\]

    • 2 years ago
  7. ash2326 Group Title
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    Yeah speed will be \[\sqrt {( v\ \cos \alpha)^2+x^2}\]

    • 2 years ago
  8. ash2326 Group Title
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    \[x^2-(u\sin \alpha)^2=2\times (-g)\times h\]

    • 2 years ago
  9. zaphod Group Title
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    |dw:1349871122014:dw|

    • 2 years ago
  10. ash2326 Group Title
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    Actually the velocity has two components v cos alpha and x Speed will be the magnitude

    • 2 years ago
  11. ash2326 Group Title
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    yes

    • 2 years ago
  12. zaphod Group Title
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    thanks :)

    • 2 years ago
  13. ash2326 Group Title
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    welcome (:

    • 2 years ago
  14. Algebraic! Group Title
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    \[speed = \sqrt{(V \sin \alpha)^{2}-2gh +(V \cos \alpha)^2}\] \[= \sqrt{(V^2 (\sin ^2 \alpha +\cos^2 \alpha)-2gh}\] \[= \sqrt{V^2 -2gh}\]

    • 2 years ago
  15. zaphod Group Title
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    thanks algebraic :)

    • 2 years ago
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