In the old West , a desperado fires a bullet into an open water tank Fig , creating a hole a distance h below the water surface . What is the speed v of the water exiting the tank ?

Explanation

From Eq.
`R_(v) = av = A v_(0)`
and thus `v_(0) = (a)/(A) "" v`
Because ` a lt lt A` , we see that `v_(0) lt lt v` . To apply Bernoulli.s equation , we take the level of the hole as our reference level for measuring elevations (and thus gravitational potential energy) . Nothing that the pressure at the top of the tank and at the bullet hole is the atmospheric pressure `p_(0)` (because both places are exposed to the atmosphere) , we write Eq. as
`p_(0) + (1)/(2) rho v_(0)^(2) + rho gh = p_(0) + (1)/(2) rho v^(2) + rho g(0) `.
(Here the top of the tank is represented by the left side of the equation and the hole by the right side . The zero on the right side indicates that the hole is at our reference level ) . Before we solve Eq. for v , we can use our result that `v_(0) lt lt v` to simplify it , We assume that `v_(0)^(2)` and thus the term `1//2 rho v_(0)^(2)` in Eq. is negligible relative to the other terms , and we drop it . Solving the remaining equation for v then yields `v = sqrt(2 gh)`
This is the same speed that an object would have when falling a height h from rest.