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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. Delete Cancel Submit

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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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• 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

• 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

• 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|

• 2 years ago
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yup! :)

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

• 2 years ago
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|dw:1354620745066:dw|

13. mayankdevnani
• 2 years ago
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then,

• 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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• 2 years ago
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you're welcome :)

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