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H/2.
let the distance be x.
then range(R)=2 . [(3H-x).x]^1/2 ......................... [ since t=(2x/g)^1/2 and v={2g(3H-x)}^1/2 ]
R^2 = 4.(3Hx-x^2)
for R to be max...R^2 should be max.
hence differentiating and putting d(R^2)/dx=0.we get 3H-2x=0
x=3H/2 (ANS).
you have to differentiate it
3h/2
use torcellies law to find the horizontal velocity then as projectile motin calcute the range
then diffrentiate to get the desired result
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