Question

please solve this one!

Answer 1

F=ma
so a=F/m

S1=1/2×a×t^2

=1/2×F/m×100

=50F/m

S2=1/2×F/m×20^2=1/2×F/m×400

So S2=200F/m

S1/S2=(50F/m)÷200F/m

S1/S2=1/4

so S2=4S1

Answer 2

Hope u like my process
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Here in the given question,

It states that force applied is same and the same particle is to be mentioned thus the mass is also constant.

Now..

By newton's 2nd law..

=> F = mass × Acceleration .

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Note:-
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=> mass and force are constant so

Acceleration = constant.

=> freely falling body states initial velocity is 0.

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Thus,

$= > S= \frac{1}{2} a {t}^{2}$

Since Acceleration is constant..

So,

=>Distance directly proportional to (time)²

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So...

$\bf = > \frac{ S_{1}}{ S_{2}} = \blue{ \frac{ {( t_{1} )}^{2} }{ {( t_{2} )}^{2} } = \frac{100}{400} } = \frac{1}{4} \\ \\ \: \: \: \: \: \bf \underline{ \green{so} }\: \: \\ \\ = > S_{1} = 4 S_{2} \: \: \: \:$
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Hence, option d(✔️) is the required answer.
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Hope this is ur required answer ❤️

Proud to help you ❤️