If the time taken by a aparticle 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.What is the inclination of the plane??
32√2h/g = √2h/gsin 2 @ [ed]
2(sin@) = 1 ...
sin @ = 1/2 .....
@ = pi / 6
11nicche 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!!!!!!!!!
11yeah 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
11it is also intuitive because the coin falls with g = gsinθ[1]
and 4 further refral #3
3i 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 ...
