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unkabogable Group Title

a stone is dropped from a captive balloon at an elevation of 1000ft,. two sec later another stone is projected vertically upward from the ground with a velocity of 248ft per sec. if g is 32ft per sec squared when and where will the stones pass each other?

  • 2 years ago
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  1. razor99 Group Title
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    wow is that a physics question

    • 2 years ago
  2. unkabogable Group Title
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    obviously..yes :)

    • 2 years ago
  3. razor99 Group Title
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    this is the first time i ever saw a physics question lol cuz i study business stream

    • 2 years ago
  4. sauravshakya Group Title
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    can u write a equation for the first stone distance travelled by it in t seconds

    • 2 years ago
  5. unkabogable Group Title
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    i know that: \[V _{01}=0 ft/s\] \[V _{02}=248 ft/s\] \[g=32.2 ft/s^2\] \[t _{1}=t _{2}+2\]

    • 2 years ago
  6. sauravshakya Group Title
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    @unkabogable can u?

    • 2 years ago
  7. sauravshakya Group Title
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    So for the first stone distance travelled by it in t seconds is = ut + 1/2 at^2 = 16t^2

    • 2 years ago
  8. sauravshakya Group Title
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    right?

    • 2 years ago
  9. unkabogable Group Title
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    yes.

    • 2 years ago
  10. unkabogable Group Title
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    yes..

    • 2 years ago
  11. sauravshakya Group Title
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    now find, for the first stone distance travelled by it in 2 seconds

    • 2 years ago
  12. sauravshakya Group Title
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    can u?

    • 2 years ago
  13. unkabogable Group Title
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    yeah...wait: so i have these eq. \[Y _{1}=1000-Y _{2}\] \[Y _{2}=V _{02}t _{2}-1/2(32.2)t _{2}^2\] then i'll substitute and i'll verify my answer later

    • 2 years ago
  14. sauravshakya Group Title
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    Let me do it then, for the first stone distance travelled by it in 2 seconds=16*2^2 = 64ft now velocity of it after two seconds = (o^2 + 2*32 *64)^1/2 = 64ft/s

    • 2 years ago
  15. sauravshakya Group Title
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    Now,Now, after two seconds, height of the stone after t seonds =1000-64-( ut+1/2 at^2) = 936-64t - 16t^2

    • 2 years ago
  16. sauravshakya Group Title
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    then for the next stone: height of the stone after t seconds = ut + 1/2 at^2 = 248t + 1/2 (-32) t^2 =248t - 16t^2

    • 2 years ago
  17. sauravshakya Group Title
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    Now, when the the stones pass each other , 936-64t - 16t^2=248t - 16t^2 936=312t t=3seconds

    • 2 years ago
  18. sauravshakya Group Title
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    so the stone will pass each other at 3+2 seconds = 5 seconds

    • 2 years ago
  19. sauravshakya Group Title
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    got it?

    • 2 years ago
  20. unkabogable Group Title
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    yes..thanks again!! :)

    • 2 years ago
  21. sauravshakya Group Title
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    WElcome

    • 2 years ago
  22. sauravshakya Group Title
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    No, can u find this where will the stones pass each other

    • 2 years ago
  23. sauravshakya Group Title
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    Now

    • 2 years ago
  24. unkabogable Group Title
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    (248)(5-2)\[(248)(5-2)- \frac{ (32.2(5-2)^2) }{ 2} = 600 ft\] ??

    • 2 years ago
  25. sauravshakya Group Title
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    Yep........ 600 ft above the ground

    • 2 years ago
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