mayankdevnani
  • mayankdevnani
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.
Physics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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
mayankdevnani
  • mayankdevnani
any short method very high level method.. @Inspired
mayankdevnani
  • mayankdevnani
@UnkleRhaukus @phi @Shadowys plz help

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anonymous
  • anonymous
that method is quite quick and high level enough. :)
mayankdevnani
  • mayankdevnani
but i am in 9 class can you tell me step by step
anonymous
  • anonymous
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|
mayankdevnani
  • mayankdevnani
i don't understand (sub it in t......)this line words are not so clear plz draw it.. @Shadowys
anonymous
  • anonymous
substitute the values u=0, and t=T/3 into S=ut+1/2at^2
mayankdevnani
  • mayankdevnani
|dw:1354620517376:dw|
anonymous
  • anonymous
yup! :)
mayankdevnani
  • mayankdevnani
next....
anonymous
  • anonymous
|dw:1354620745066:dw|
mayankdevnani
  • mayankdevnani
then,
anonymous
  • anonymous
then, since when T/3 --> h/9 , then at that time, the object is h - h/9 above the ground, right?|dw:1354620913980:dw|
mayankdevnani
  • mayankdevnani
ohkkk!! thnx.. @Shadowys
anonymous
  • anonymous
you're welcome :)

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