The popular Super Soaker water guns are powered
by compressed air (provided by a hand pump used by
the operator.) A certain model has a cylindrical
reservoir (where the water is held) 8.00 cm in diameter
and a barrel (where the water leaves the gun) 1.00 cm
in diameter. If aimed at the optimum angle, water
leaving the barrel can hit a target 14.0 m away. If the
end of the barrel is 30.0 cm higher than the reservoir,
find:
(a) the flow rate of the water leaving the barrel;
(b) the gauge pressure of the water in the reservoir.
(Assume air resistance acting on the water in flight is
negligble.)
To find the flow rate of the water leaving the barrel, we can use the principle of conservation of energy.
(a) The flow rate can be determined by calculating the velocity of the water leaving the barrel. We can use the equation of motion for vertical motion under constant acceleration:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
y is the vertical distance traveled (30.0 cm or 0.3 m),
y0 is the initial vertical position (the height of the reservoir),
v0y is the initial vertical velocity,
t is the time of flight, and
g is the acceleration due to gravity.
Since the water leaves the barrel horizontally, the initial vertical velocity v0y is zero, and the equation becomes:
y = - (1/2) * g * t^2
Simplifying for t:
t = sqrt(2 * y / g)
Substituting the values:
t = sqrt(2 * 0.3 m / 9.8 m/s^2) ≈ 0.243 s
Since we know that the horizontal distance traveled by the water is 14.0 m and the time of flight is 0.243 s, we can calculate the velocity v of the water leaving the barrel using the equation:
v = d / t
v = 14.0 m / 0.243 s ≈ 57.6 m/s
The flow rate Q is given by the equation:
Q = A * v
Where:
Q is the flow rate,
A is the cross-sectional area of the barrel,
and v is the velocity of the water.
The cross-sectional area of the barrel can be calculated using the formula for the area of a circle:
A = π * r^2
Where r is the radius of the barrel, which is half of the diameter.
Given the diameter of the barrel is 1.00 cm, the radius is 0.5 * 1.00 cm = 0.5 cm or 0.005 m.
Substituting the values:
A = π * (0.005 m)^2
A ≈ 0.0000785 m^2
Finally, substituting the values into the flow rate equation:
Q = 0.0000785 m^2 * 57.6 m/s ≈ 0.00451 m^3/s
Therefore, the flow rate of the water leaving the barrel is approximately 0.00451 m^3/s.
(b) To find the gauge pressure of the water in the reservoir, we can use the principle of fluid statics.
The pressure at any depth in a fluid is given by:
P = P0 + ρgh
Where:
P is the pressure at a given depth,
P0 is the atmospheric pressure,
ρ is the density of the fluid,
g is the acceleration due to gravity,
and h is the depth.
Since we need to find the gauge pressure (the pressure above atmospheric pressure), we can ignore P0.
Assuming the water reservoir is filled only with water, the density of water is approximately ρ = 1000 kg/m^3.
Given the depth h is 30.0 cm or 0.3 m, and g is 9.8 m/s^2, we can calculate the gauge pressure P:
P = ρgh
P = 1000 kg/m^3 * 9.8 m/s^2 * 0.3 m
P ≈ 2940 Pa
Therefore, the gauge pressure of the water in the reservoir is approximately 2940 Pa.