In the figure shown find the value of `h_(2)` for maximum range `R` if

(a) `h_(1)=4H` (b)`h_(1)=8H`
Also find the value of this maximum range in both cases.

Answer

A, B, C

Explanation

(a) `h_(1)+6H=4H+6H=10H=h`(say)
Maximum range will be obtained from
`h_(2)=(h)/(2)=5H`.
and this maximum range will be,
`R_(max)=h=10H`.
(b) In this case,
`h=h_(1)+(6H)=(8H)+(6H)=14H`
`(h)/(2)` or `7H` point lies on the table. so, maximum range will be obtain from the bottommost point of the liquid container or `h_(2)=6H`
and this maximum range will be `R=2 sqrt(h_(2)(h-h_(2))`
`=2 sqrt(6Hxx8H)=8 sqrt(3) H`.