In: Physics
A firefighter carries a hose up a ladder to a height of 10.0 m, so that she can spray water onto a burning roof that is 9.00 m high. She holds the hose horizontally, and sees that the water strikes the roof 7.0 m in front of her position (i.e. horizontal component of displacement is 7.0 m). The hose is connected to a very large, pressurized chamber in the fire truck, which is sitting at ground level. What is the pressure in the large reservoir? (Ignore air resistance; don’t be off by 1 atm.)
*include sketch/diagram
let v is the velocity of the water with which they come out of the hose.
let t is the time taken by the water to hit the roof.
now use, y = (1/2)*g*t^2
10 - 9 = (1/2)*9.8*t^2
==> t = sqrt(2/9.8)
= 0.4517 s
now use, v = x/t
= 7/0.4517
= 15.5 m/s
let v2 = v = 15.5 m/s (at hose)
P2 = 1 atm = 1.013*10^5 pa (at hose)
h2 = 10 m
v1 = 0 (inside chamber)
h1 = 0
P1 = ?
now use bernoulli's equation P1 + (1/2)*rho*v1^2 + rho*g*h1 = P2 + (1/2)*rho*v2^2 + rho*g*h2
P1 + 0 + 0 = P2 + (1/2)*rho*v2^2 + rho*g*h2
P1 = P2 + (1/2)*rho*v2^2 + rho*g*h2
= 1.013*10^5 + (1/2)*1000*15.5^2 + 1000*9.8*10
= 3.19*10^5 pa or 3.15 atm p