A) x is equal, if depth = y or H - y
B) x is maximum for \[\operatorname{y}=\frac{H}{2}\]
C) a & b correct
D) None
Correct Answer: C
Solution :
\[\operatorname{At} y =\frac{H}{2} maximum range is obtained\] which is equal to H \[{{R}_{h}}={{R}_{\operatorname{H}-h}} Velocity of efflux = \sqrt{2gy}\] \[\operatorname{H} - y = \frac{1}{2}g{{T}^{2}} uy = 0 for horizontal motion\] \[\operatorname{T}=\sqrt{\frac{2\left( H-y \right)}{\operatorname{g}}}\] \[\operatorname{Range}=\sqrt{\frac{2gy\times 2\left( H-y \right)}{g}}=2\sqrt{\left( H-y \right)\operatorname{y}}\] \[\operatorname{For} x to be maximum \frac{dx}{dy}=0\] \[\operatorname{We} get\,\,y=\frac{H}{2}\] \[\operatorname{Range} is equal for \to y = \frac{H}{2} or\] \[\operatorname{H}-y\Rightarrow -\frac{H}{2}=\frac{H}{2}\]You need to login to perform this action.
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