Water is filled in a tank up to a height of 3 | vedclass

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(D) Let the total height of the water level from the ground be $H = 3 \,m + 1 \,m = 4 \,m$. Let the height of the hole from the ground be $y$. The dep

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$2 \,m$ from ground

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(D) Let the total height of the water level from the ground be $H = 3 \,m + 1 \,m = 4 \,m$. Let the height of the hole from the ground be $y$. The depth of the hole from the free surface of the water is $h = H - y = 4 - y$. The velocity of efflux is $v = \sqrt{2gh}$. The time taken for the water to reach the ground is $t = \sqrt{\frac{2y}{g}}$. The horizontal range $R$ is given by $R = v \cdot t = \sqrt{2g(4-y)} \cdot \sqrt{\frac{2y}{g}} = 2\sqrt{y(4-y)}$. For maximum range,the term inside the square root,$f(y) = 4y - y^2$,must be maximum. Taking the derivative with respect to $y$ and setting it to zero: $\frac{df}{dy} = 4 - 2y = 0$,which gives $y = 2 \,m$. Thus,the hole should be at a height of $2 \,m$ from the ground.

Water is filled in a tank up to a height of $3 \,m$. The base of the tank is at a height of $1 \,m$ above the ground. What should be the height of a hole made in the side of the tank so that water can be sprayed up to the maximum horizontal distance on the ground?

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