**Q. Evaluate Height of Water in Tank?**

A water tank has spiral a small leak at a point 2 feet from its base on the ground. A second leak sprightly over the first is 5 feet from the base.

A passing physics student become aware of that the two streams issuing from the tank were striking the ground at the same spot. He after that realized he could calculate the height of the water in the tank. What were his end results?

We start by determining the velocity of the water issuing from the two leaks using Torricelli's theorem. We will utilize subscripts to link the relevant equations to their respective streams and insert the known values at the end.

v_{1}^{2} = 2g(h - h_{1})

v_{2}^{2}= 2g(h - h_{2})

We are able to find the time required for each stream to strike the ground from the kinematics equation y = vt_{0} + 1/2gt^{2}. For this numerical problem t_{0 }= 0.

h_{1} =1/2gt_{1}^{2}

h_{2 }=1/2gt_{2}^{2}

Solving

t_{1}^{2} =2h_{1}/g

t_{2}^{2}_{ }=2h_{2}/g

The horizontal distance travelled by every stream is vt. We have v_{1}t_{1} = v_{2}t_{2}

Or

(v_{1}t_{1})^{2} = (v_{2}t_{2})^{2}

Thus substituting from the above

(2g(h - h_{1}))2h_{1}/g= (2g(h - h_{2}))2h_{2}/g

(h - h_{1})h_{1} = (h - h_{2})h_{2}

hh_{1} - h_{1}^{2} = hh_{2} - h_{2}^{2}

h(h_{1 }- h_{2}) = h_{1}^{2} - h_{2}^{2}

h =((h_{1}^{2}- h_{2}^{2})/(h_{1 }- h_{2}))=((h_{1} + h_{2})(h_{1} - h_{2}))/h_{2} - h_{2 } thus,

h = h_{1 }+ h_{2}.

Specified that h_{1} = 2 ft and h_{2} = 5 ft the height of the water in the tank h = 7 ft.