Question

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solve it as fast as u cn.....

Answer 1

Answer:

Let the height of tower be 'x'.

Given:

u = 0m/s

Time to cover:

$\implies$ $\sf{\frac{5}{9x}}$ of building be t sec.

$\implies$ $\sf{\frac{4x}{9}}$ of the building be t - 1 sec.

We know that:

$\boxed{\sf{S_{t} = u + \frac{1}{2} g(2t - 1)}}$

$\sf{\frac{5x}{9} = 5(2t - 1)}$ - - - - (1)

We know that:

$\boxed{\sf{S = ut + \frac{1}{2}gt^{2}}}$

$\sf{\frac{4x}{9} = 0 + 5(t - 1)^{2}}$ - - - - (2)

Dividing (1) and (2),

We get:

$\implies$ $\sf{ \frac{ \frac{5x}{9} }{ \frac{4x}{9} } = \frac{5(2t - 1)}{5(t - 1) ^{2}} }$

$\implies$ $\sf{5t^{2}+5-10t=8t-4}$

$\implies$ $\sf{5t^{2} -18t +9 =0}$

Solving 't' we get: t = 3 sec

Note: Check this attachment.

Therefore:

The time of fall is 3 sec.

Answer 2

Answer:

Answer:

Let the height of tower be 'x'.

Given:

u = 0m/s

Time to cover:

\implies⟹ \sf{\frac{5}{9x}}

9x

5

of building be t sec.

\implies⟹ \sf{\frac{4x}{9}}

9

4x

of the building be t - 1 sec.

We know that:

\boxed{\sf{S_{t} = u + \frac{1}{2} g(2t - 1)}}

S

t

=u+

2

1

g(2t−1)

\sf{\frac{5x}{9} = 5(2t - 1)}

9

5x

=5(2t−1) - - - - (1)

We know that:

\boxed{\sf{S = ut + \frac{1}{2}gt^{2}}}

S=ut+

2gt 2

\sf{\frac{4x}{9} = 0 + 5(t - 1)^{2}} 94x

=0+5(t−1) 2

- - - - (2)

Dividing (1) and (2),

We get:

\implies⟹ \sf{ \frac{ \frac{5x}{9} }{ \frac{4x}{9} } = \frac{5(2t - 1)}{5(t - 1) ^{2}} }

94x

95x

=

5(t−1) 2

5(2t−1)

\implies⟹ \sf{5t^{2}+5-10t=8t-4}5t

2+5−10t=8t−4

\implies⟹ \sf{5t^{2} -18t +9 =0}5t

2 −18t+9=0

Solving 't' we get: t = 3 sec

Note: Check this attachment.

Therefore:

The time of fall is 3 sec.