A stream of water from a fire-fighter\'s hose hits a vertical wall. The firefigh

Question

A stream of water from a fire-fighter's hose hits a vertical wall. The firefighter knows that 16.9 kg of water are emitted from the hose per second, and that the water has a speed of 45.5 m/s when it hits the wall. At the instant when it hits the wall, the direction of the velocity vector of the water is 25.0 degrees BELOW the horizontal. Calculate the average force exerted by the water on the wall, assuming that the vertical component of the velocity vector of the water is unchanged but the horizontal component of the velocity vector is reversed when the water rebounds from the wall.

Explanation / Answer

mass of water emitting from the hose per second m/t = 16.9 kg

angle theta is 25 degrees below the horizontal, speed v = 45.5 m/s

force = rate of change of momentum

now water horizontal velocity is Vh = V cos theta = 45.5*cos25= 41.24 m/s
change in velocity for velocity reverse upon striking the wall is = 2*vh = 2*41.24 =82.48 m/s

   now force = rate of change of momentum
  
       F= m Dv/t = m/t*(Dv)
           = 16.9(82.48)
           = 1393.912 N
           F = 1393.912 N

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