A body of mass m is proejected horizontally from the cliff of height h with a velocity v.It touches the ground at a distance of 250mfrm the bottom of a cliff.If a body of mass 2m is projected horizontlly with a velcity v/2 from cliff of height 4h,it touches the ground at a distance of

h=Uyt+1/2 at^2

Uy=0

 so t=sqrt(2h/a)=sqrt(2h/g)

horizontal distance S =Vx.t

                             S =sqrt(2h/g)V..............1

 so S1/S2 =sqrt(h1/h2). V1/V2..................2

   in first case S=S1=250 , velocity V1=v and height h1=h

   in second case S=S2=? ,velocity V2=v/2 and height H2=4h

 sustituting these values in eq 2

we get S2=250m

Approved

the answer is 125 meters.because the horizontal displacement does not depend on the mass of the projectile .it only depends on initial velocity..

Approved

the answer is 250 rite..