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

ALL THE BEST!