32√2h/g = √2h/gsin 2 @ [ed]
2(sin@) = 1 ...
sin @ = 1/2 .....
@ = pi / 6
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??
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8 Answers
3
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 ...