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Particle is dropped from the height of 20 m from the horizontal ground. There is wind blowing due to which horizontal acc. of the particle becomes 6 m/s/s. Find the horizontal disp. of the particle till it reaches the ground.
Height = 20 m
so by using the following equation for motion in vertical direction
we know g=10 and u=0 and s= 20
s= ut + 1/2 gt^2
20=1/2*10*t^2; (as u=0)
thus t= 2 sec
Hence the particle will reach the ground after 2 sec
a(in horizontal direction)=6
Again usng the same equation but now for horizontal motion of particle
s=ut + 1/2 at^2
we know t=2 and u=0
so s=1/2*6*4
s=12 m
So horizontal displACEMENT OF PARTICLE =2M
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