A tank is filled with water up to a height $H$. Water is allowed to come out of a hole $ P$ in one of the walls at a depth $ D $ below the surface of water. Express the horizontal distance $x$ in terms of $ H $ and $D$
Diffcult
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(c)Time taken by water to reach the bottom
= $t = \sqrt {\frac{{2(H - D)}}{g}} $
and velocity of water coming out of hole, $v = \sqrt {2gD} $
Horizontal distance covered $x = v \times t$
= $\sqrt {2gD} \times \sqrt {\frac{{2(H - D)}}{g}} $= $2\sqrt {D(H - D)} $
= $t = \sqrt {\frac{{2(H - D)}}{g}} $
and velocity of water coming out of hole, $v = \sqrt {2gD} $
Horizontal distance covered $x = v \times t$
= $\sqrt {2gD} \times \sqrt {\frac{{2(H - D)}}{g}} $= $2\sqrt {D(H - D)} $
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