Water stands at a depth H in a tank whose side wails are vertical. A hole is made on one of the walls at a depth h below the water surface. Find at what distance from the foot of the wall does the emerging stream of water strike the floor?

   x = 2h

Explanation:

This is an exercise in fluid mechanics, let's find the speed of the water in the hole

       P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

where the subscript 1 is for the tank surface and the subscript 2 is for the depth of the hole

the pressure inside and outside the tank is the same

       P₁ = P₂

we must measure the distance from the same reference point, let's locate it on the surface of the water

       y₁ = 0

       y₂ = h

Suppose the gap is small compared to the diameter of the tank

        v₁ «v₂

       v₂² = 2 g (0-h)

This is the speed of the outlet water in the tank, as the force is horizontal this speed is horizontal.

Let's use the projectile launch ratios

         vₓ =

         y = y₀ + v₀ t - ½ g t²

the height when reaching the floor is y = 0,

the initial height is measured from the floor therefore y₀ = H-h

        0 = (H-h) + 0 - ½ g t²

        t =

we look for the distance x traveled

        x = vₓ t

        x =

        x =

        x = 2h

a and c

Explanation: i took a guess but let me know if i'm correct.