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(A) Let $v_1$ be the velocity of water at the hole (exit velocity) and $v_2$ be the velocity of water when it hits the ground.Using Torricelli's Law,t

Vedclass · Accepted Answer

$x = r\left( \frac{H}{H + h} \right)^{\frac{1}{4}}$

Vedclass · Answer

(A) Let $v_1$ be the velocity of water at the hole (exit velocity) and $v_2$ be the velocity of water when it hits the ground.Using Torricelli's Law,the velocity at the hole is $v_1 = \sqrt{2gH}$.Using the equation of motion for a freely falling body,the velocity at the ground is $v_2 = \sqrt{v_1^2 + 2gh} = \sqrt{2gH + 2gh} = \sqrt{2g(H + h)}$.According to the equation of continuity,the volume flow rate remains constant:$A_1 v_1 = A_2 v_2$$\pi r^2 v_1 = \pi x^2 v_2$$r^2 \sqrt{2gH} = x^2 \sqrt{2g(H + h)}$Squaring both sides:$r^4 (2gH) = x^4 (2g(H + h))$$x^4 = r^4 \frac{H}{H + h}$$x = r \left( \frac{H}{H + h} \right)^{\frac{1}{4}}$

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