Question

Please don't post irrelevant answers
answer atleast 3 questions ​

Answer 1

1. By second equation of motion, displacement of the dropped body,

$\sf{\longrightarrow s=ut+\dfrac{1}{2}\,gt^2}$

Initial speed is zero since the body is dropped from height.

$\sf{\longrightarrow s=\dfrac{1}{2}\,gt^2}$

This implies,

$\sf{\longrightarrow s\propto t^2}$

$\sf{\longrightarrow t\propto\sqrt s}$

Therefore,

$\sf{\longrightarrow \dfrac{t'}{t}=\sqrt{\dfrac{\left(\dfrac{h}{2}\right)}{h}}}$

$\sf{\longrightarrow\underline{\underline{t'=\dfrac{t}{\sqrt2}}}}$

2. The displacement of the ball dropped freely is,

$\sf{\longrightarrow s_1=\dfrac{1}{2}\,gt^2}$

And that of the ball thrown vertically downwards.

$\sf{\longrightarrow s_2=vt+\dfrac{1}{2}\,gt^2}$

Then the separation between them is,

$\sf{\longrightarrow \Delta s=s_2-s_1}$

$\sf{\longrightarrow \Delta s=vt+\dfrac{1}{2}\,gt^2-\dfrac{1}{2}\,gt^2}$

$\sf{\longrightarrow\underline{\underline{\Delta s=vt}}}$

3. The distance moved by a freely falling body in $\sf{n^{th}}$ second is the change in positions of the body after $\sf{n}$ and $\sf{n-1}$ seconds of dropping, i.e.,

$\sf{\longrightarrow \Delta s_n=s_n-s_{n-1}}$

$\sf{\longrightarrow \Delta s_n=\left(un+\dfrac{1}{2}\,gn^2\right)-\left(u(n-1)+\dfrac{1}{2}\,g(n-1)^2\right)}$

$\sf{\longrightarrow \Delta s_n=u+\dfrac{1}{2}\,g(2n-1)}$

For a freely falling body, $\sf{u=0.}$

$\sf{\longrightarrow \Delta s_n=\dfrac{1}{2}\,g(2n-1)}$

This implies,

$\sf{\longrightarrow \Delta s_n\propto2n-1}$

Therefore,

$\sf{\longrightarrow \Delta s_1:\Delta s_2:\Delta s_3:\,\dots\,:\Delta s_n=(2\cdot1-1):(2\cdot2-1):(2\cdot3-1):\,\dots\,:(2n-1)}$

$\sf{\longrightarrow\underline{\underline{\Delta s_1:\Delta s_2:\Delta s_3:\,\dots\,:\Delta s_n=1:3:5:\,\dots\,:(2n-1)}}}$

4. By third equation of motion, displacement of the body dropped is given by,

$\sf{\longrightarrow v^2=2gh}$

Since initial velocity is zero for a freely falling body.

Then,

$\sf{\longrightarrow h=\dfrac{v^2}{2g}}$

Taking $\sf{g=10\ m\,s^{-2},}$

$\sf{\longrightarrow h=\dfrac{20^2}{2\cdot 10}}$

$\sf{\longrightarrow\underline{\underline{h=20\ m}}}$

Answer 2

Explanation:

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