kinematics

2√2h/g = √2h/gsin 2 @ [ed]

2(sin@) = 1 ...

sin @ = 1/2 .....

@ = pi / 6

nicche gravity g sinθ lagti hai
and u need

time taken by a a particle sliding down a smooth inclined plane is twice of that it would take if it would have fallen down through the vertical height of the inclined plane

iska matlab
in a free fall

t=√2h/g
so replace g here by g sinθ and put the conditions given in the question!!!!!!!!!

answer pie/6 hai bhai

yeah answer is pie/6

time taken along inclined plane =1/sinθ√2h/g

time during free fall=√2h/g

using these sinθ=1/2

so pie/6

it is also intuitive because the coin falls with g = gsinθ[1]
and 4 further refral #3

i took rong comp. yerp its gsin@cos(90-@) = gsin^2@........[hav edited abve]

thatz the vertical component of accn.

and manipal ur #4 is rong if ur puttin gsin@ u have to take dist. as h/sin@ not just h ...

thx ppl