Question

please solve this problem!

Answer 1

Hope u like my process
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=> Let the total time taken be t sec.

=> Let the total height be x m.

Since body is falling freely,

=> Initial velocity (u) =0

=> Acceleration=gravity =10 m/s (constant).

We know,
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$\bf = > s = ut + \frac{1}{2} g {t}^{2} \\ \\ \underline{ \green{since \: \: u = 0 \: \: and \: \: g = constant}}$
So,..

Distance is directly proportional to (time)².
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=> Distance travelled in last second= 0.36x m.

=> Distance travelled in (t-1)seconds

= (x - 0.36x) = 0.64 m

So...We Can say.. That..,

$= > \bold{ \frac{ \blue{x}}{ \blue{0.64x} }} = \frac{ {( \blue{t})}^{2} }{ {( \blue{t - 1})}^{2} } \\ \\ = > { \it \: \frac{ \blue{1}}{ \blue{0.64}} } = \frac{ \blue{ {t}^{2} }}{ { \blue{(t - 1)^{2}} }} \\ \\ = > \frac{ \blue{1}}{ \blue{0.8}} = \frac{ \blue{t}}{ \blue{t - 1}} \\ \\ = > \blue{ \bf{t - 1}} = \blue{ \bold0.8t } \\ \\ = > \blue{t - 0.8t }= \blue{ 0.2t }= \blue{1} \\ \\ = > t = \frac{ \blue{1}}{ \blue{0.2}} = \underline{ \blue{5 \: \: sec}}$

So...total height =½× g×t² =½× 10×5²

$= \bf \underline{ \orange{125 \: m}}$

Hence, option d(✔️) is your required height
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Hope this is ur required answer

Proud to help you

Answer 2

Let H be the height of the building and t be the time to fall through this height H.  

Then H = ½ g t^2 = 5 t^2. ----------------- (1)  

In (t-1) second the distance traveled is 1 - 0.36 H = 0.64 H.  

0.64 H = 5 (t-1)^2.---------------------------(2)  

(2) / (1) gives 0.64 = (t-1)^2/ t^2  

Or 0.8 = (t-1) / t  

Solving we get t= 5 second.  

Using (1) we get H = 125 m.

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