Question

answer the above attachment.... ​

Answer 1

Answer:

$\huge\underline\pink{\tt Hey\:Mate}$

$\huge\underline\blue{\tt Ur\:Answer\:Is\:(2)41\:%}$

When a stone is dropped from a height h it hits the ground with a momentum.

P= m ✓2gh -----(1)

where

m= Mass of stone

When same stone is dropped from a height 2h (i.e 100% of initial ) then its momentum with which it hits ground becomes .

P' = m ✓2g(2h) => ✓2P -------(2)

% change in momentum = P'-P/ P× 100%

= √2P - P/P×100 %

= 41%

Hope it helps uh!!

Answer 2

Answer:

$\huge\underline\orange{\mathcal Answer}$

$\huge{\boxed{\tt (2)41%}}$

When stone is droped from height h :-

$\large{\mathcal P = m\sqrt{2gh}----->(1)}$

When stone is droped from height 2h :-

$\large{\mathcal P' =m\sqrt{2g(2h)}}$

$\large{\mathcal P'=\sqrt{2}P----->(2)}$

$\large{\boxed{\tt %∆P={\frac{P-P'}{P}×100%}}}$

@14riyaz14