Let's break down this problem step by step. First, we need to understand the concept of projectile motion and how it relates to the flow of water from the hole in the tank. The stream of water coming out of the hole can be treated as a projectile, with an initial velocity equal to the velocity of the water at the hole's depth.

Step 1: Finding the Horizontal Distance x

To find the horizontal distance x from the foot of the wall at which the stream strikes the floor, we can use the equation for projectile motion: x = (v^2 * sin(2θ)) / g, where v is the velocity of the water at the hole's depth, θ is the angle of projection, and g is the acceleration due to gravity.

Given that the water is flowing vertically downward from a height H, the initial velocity v can be found using the equation v = sqrt(2gH).

Substitute the value of v into the equation for x to get x = 2 * sqrt(H(h - h)), which simplifies to x = 2 * sqrt(Hh).

Step 2: Finding the Depth for Same Range

To determine if a hole could be punched at another depth so that the second stream would have the same range, we need to find a depth h' that satisfies the equation x = 2 * sqrt(Hh').

Squaring both sides of the equation gives Hh' = Hh, which simplifies to h' = h. Therefore, a hole could be punched at the same depth h to achieve the same range.

Step 3: Maximum Distance from Base of the Tank

To find the depth at which the hole should be placed to make the emerging stream strike the ground at the maximum distance from the base of the tank, we need to maximize the horizontal distance x = 2 * sqrt(Hh).

Take the derivative of x with respect to h and set it equal to zero to find the critical point: dx/dh = 0. Solving for h gives h = H/4.

Substitute h = H/4 back into the equation for x to find the maximum distance: x = 2 * sqrt(H(H/4)) = H * sqrt(2).

Therefore, the hole should be placed at a depth of H/4 to make the emerging stream strike the ground at the maximum distance of H * sqrt(2) from the base of the tank.