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mayankdevnani

  • 2 years ago

A ball is released from the top of a tower of height 'h' meters. It takes 'T' seconds to reach the ground.What is the position of the ball in T/3 seconds? a) h/9 metres from ground. b)7h/9 m from ground. c) 8h/9 m from ground. d) 17h/18 m from ground.

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  1. Inspired
    • 2 years ago
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    ok, so using the position function, you see that \[\Delta y = .5 g t^2 \] and \[\Delta y = .5g (t/3)^2 = (.5 g t^2) / 9\] when t =t/3 so essentially, you have, h-h/9= 8h/9

  2. mayankdevnani
    • 2 years ago
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    any short method very high level method.. @Inspired

  3. mayankdevnani
    • 2 years ago
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    @UnkleRhaukus @phi @Shadowys plz help

  4. Shadowys
    • 2 years ago
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    that method is quite quick and high level enough. :)

  5. mayankdevnani
    • 2 years ago
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    but i am in 9 class can you tell me step by step

  6. Shadowys
    • 2 years ago
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    the motion formula is \(S=ut + \frac{1}{2} at^2\), as u=0, and a=g, it simplifies to \(h=\frac{1}{2} aT^2\) when u=0, t=T/3, sub it in to get \(S= \frac{1}{2} a\frac{T^2}{9}=\frac{\frac{1}{2} aT^2}{9}=h/9\) subtracting the remaining height givens 8h/9|dw:1354619702849:dw|

  7. mayankdevnani
    • 2 years ago
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    i don't understand (sub it in t......)this line words are not so clear plz draw it.. @Shadowys

  8. Shadowys
    • 2 years ago
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    substitute the values u=0, and t=T/3 into S=ut+1/2at^2

  9. mayankdevnani
    • 2 years ago
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    |dw:1354620517376:dw|

  10. Shadowys
    • 2 years ago
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    yup! :)

  11. mayankdevnani
    • 2 years ago
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    next....

  12. Shadowys
    • 2 years ago
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    |dw:1354620745066:dw|

  13. mayankdevnani
    • 2 years ago
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    then,

  14. Shadowys
    • 2 years ago
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    then, since when T/3 --> h/9 , then at that time, the object is h - h/9 above the ground, right?|dw:1354620913980:dw|

  15. mayankdevnani
    • 2 years ago
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    ohkkk!! thnx.. @Shadowys

  16. Shadowys
    • 2 years ago
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    you're welcome :)

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