Physics, asked by shinjan18, 13 days ago

explain with answer both questions ​

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Answered by XxitzKing02xX
4

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1).

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A particle is dropped from the top of a tower. During its motion it covers 25/9 part of height of tower the last 1 seconds. Then find the height of tower?

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s =  \frac{1}{2}  \times a \times  {t}^{2}  =  \frac{1}{2}  {gt}^{2}

⇝x =  \frac{1}{2} \: {gt}^{2} .....(1)

⇝\frac{16x}{25}  =  \frac{1}{2} {g(t - 1)}^{2} .........(2)

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 \frac{16}{25}  =   \frac{(t - 1)} {t}^{2}

  {16t}^{2}  =  25{(t - 1)}^{2} .......(3)

WHICH GIVES T=5 SEC

(TIME CANNOT BE NEAGITIVE)

hence \: height \: x =  \frac{1}{2} \times 9.8 \times ( {5}^{2})

 = 122.5m

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2)

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Two tall building face each other and are at a distance of 180 m from each other. With what velocity must a ball be thrown horizontally from a window 55 m above the ground in one building, so that it enters a window 10.9 m above the ground in the second building?

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Height = h= 55-10.9 = 44.1

Distance= s= 180 m

time taken = t

so, h=1/2 gt²

or t= √2h/g

√2 x 44.1/9.8

=√88.2/9.8

=√9 = 3

 

S= velocity x t

180= velocity x 3

velocity =180/3 = 60 m/s

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Hope it helps you❤️

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