Question

please solve this!!!

Answer 1

For 1st 3 sec put h=1/2g( n^2) then for the next (2n-1)/2*g

Answer 2

Hope u like my process
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Formula to be used
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=> Distance travelled for time (t)

$= \green{ \: \: ut \: \: + \frac{1}{2} g {t}^{2} }$
=> Distance travelled in nth second

$= \green{u + \frac{g}{2} }( \green{2n - 1})$
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Given,
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Stone is falling freely, so,

=> Initial velocity (u) = 0

=> gravity (g) = 10 m/s

=> Distance travelled in first 3 seconds

$= ut + \frac{1}{2} g {t}^{2} \\ \\ = 0 + \frac{1}{2} \times 10 \times {3}^{2} \\ \\ = \underline{ \blue{45 \: \: m}}$

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=> Let the last second be nth second.

So,.

$= > s = u + \frac{g}{2} (2n - 1) \\ \\ = > \blue{45} = \blue{0 + \frac{10}{2}(2n - 1) } \\ \\ = > \blue{\frac{45}{5} } = \blue{2n - 1} \\ \\ = > \blue{2n} = \blue{9 + 1} = \blue{10} \\ \\ \: \: \: \underline{ \orange{ \bold {\: so} \: }} \\ \\ = > \boxed{n = \underline{ \bold{ \blue{5 \: \: sec}}}}$

So,

The last second was 5th second.
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Thus

the required answer is option b(✔️).

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Hope this is ur required answer❤️

Proud to help you ❤️