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  • A particle is projected from a point on the surface of a smooth inclined plane simultaneously another particle q is released on

IMG_20190123_090703.jpg A particle is projected from a point on the surface of a smooth inclined plane simultaneously another particle q is released on the smooth inclined plane from the same position p and q polite after t is equal to 4 seconds the speed of projection of P is

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best_answer

@Arvind gupta 

For Q

U=0,  a=gsin\Theta

So s_q=\frac{1}{2}gsin\Theta\ \ t^2=8gsin\Theta

Now for P

if the angle of projection of P with the incline surface is α 
=> ucosα = speed along the plane, a = gsinθ° = acceleration along the plane 

Similarly usinα = speed perpendicular  to the plane, a = gcosθ° = acceleration perpendicular to the plane 

time of flight (For P) = 2usinα/gcosθ° = 4 s 

usinα = 2gcosθ°

distance travelled down incline = s = ut + ½at² 
Sp= 4ucosα + 8gsinθ° 

So Sp=Sq

= 4ucosα + 8gsinθ° = 8gsinθ° 
=> cosα = 0 or α = 90°

So we get u = 2gcosθ°=

u=2gcos\Theta =2g\frac{1}{2}=g=10m/s^2

g stn6 gsin6 t2 = 8gsin6 2 2gcos6 = g = 10m/s2 2
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