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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. |
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Answer» Correct Answer - A::B::C (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`. |
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