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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
the answer is 125 meters.because the horizontal displacement does not depend on the mass of the projectile .it only depends on initial velocity..
the answer is 250 rite..
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