A firefighter mounts the nozzle of his fire hose a distance 36.9 m away from the edge of a burning building so that it sprays from ground level at a 45° angle above the horizontal. After quenching a hotspot at a height of 8.85 m, the firefighter adjusts the nozzle diameter so that the water hits the building at a height of 17.9 m. By what factor was the nozzle diameter changed? Assume that the diameter of the hose stays the same, and treat the water as an ideal fluid.
Answer:
diameter is decreased by factor 0.91
Explanation:
As we know by the equation of trajectory
\(y = xtan\theta - \frac{gx^2}{2v^2cos^2\theta}\)
here as per first given situation we know that
x = 36.9 m
y = 8.85 m
\(\theta = 45^0\)
now from above equation we have
\(8.85 = 36.9 tan45 - \frac{(9.8)(36.9)^2}{2(v^2)cos^245}\)
\(8.85 = 36.9 - \frac{13343.8}{v^2}\)
\(\frac{13343.8}{v^2} = 28.05\)
\(v = 21.8 m/s\)
now similarly after nozzle is adjusted we have
y = 17.9 m
x = 36.9 m
\(\theta = 45^0\)
now again from equation we have
\(17.9 = 36.9 tan45 - \frac{(9.8)(36.9)^2}{2(v'^2)cos^245}\)
\(17.9 = 36.9 - \frac{13343.8}{v'^2}\)
\(\frac{13343.8}{v'^2} = 19\)
\(v' = 26.5 m/s\)
Now by equation of continuity we can find the change in diameter
as we know that
\(A_1v_1 = A_2v_2\)
now we have
\(\pi d_1^2 v_1 = \pi d_2^2 v_2\)
\(d_1^2 (21.8) = d_2^2(26.5)\)
\(\frac{d_1}{d_2} = \sqrt{\frac{26.5}{21.8}}\)
\(\frac{d_1}{d_2} = 1.10\)
so we have
\(\frac{d_2}{d_1} = 0.91\)
so diameter is decreased by factor of 0.91
The nozzle diameter changed by a factor of 0.907
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Acceleration is rate of change of velocity.
\(\large {\boxed {a = \frac{v - u}{t} } }\)
\(\large {\boxed {d = \frac{v + u}{2}~t } }\)
a = acceleration (m / s²)v = final velocity (m / s)
u = initial velocity (m / s)
t = time taken (s)
d = distance (m)
Let us now tackle the problem!
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Given:
horizontal distance = x = 36.9 m
angle of projection = θ = 45°
initial height = y₁ = 8.85 m
final height = y₂ = 17.9 m
Asked:
ratio of nozzle diameter = d₂ : d₁ = ?
Solution:
The motion of the water is a parabolic motion.
Firstly, we will calculate the time taken for the water to reach the hotspot:
\(x = (u \cos \theta) t\)
\(t = x \div ( u \cos \theta )\)
\(t = x \div ( u \cos 45^o )\)
\(\boxed {t = \frac{\sqrt{2}x}{u}}\)
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Next , we could calculate the initial speed (u) of the water as it leaves the nozzle:
\(y = (u \sin \theta) t - \frac{1}{2}gt^2\)
\(y = (u \sin 45^o)( \frac{\sqrt{2}x}{u} ) - \frac{1}{2}g ( \frac{\sqrt{2}x}{u} )^2\)
\(y = x - \frac{gx^2}{u^2}\)
\(\frac{gx^2}{u^2} = x - y\)
\(u^2 = \frac{gx^2}{x - y }\)
\(u = \sqrt{ \frac{gx^2}{x - y } }\)
\(\boxed {u = x \sqrt{ \frac{g}{x - y} }}\)
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Finally , we could find the ratio of the diameter by using Continuity Equation as follows:
\(u_1 A_1 = u_2 A_2\)
\(u_1 \frac{1}{4} \pi (d_1)^2 = u_2 \frac{1}{4} \pi (d_2)^2\)
\((d_2)^2 : (d_1)^2 = u_1 : u_2\)
\((d_2)^2 : (d_1)^2 = x \sqrt{ \frac{g}{x - y_1} } : x \sqrt{ \frac{g}{x - y_2} }\)
\((d_2)^2 : (d_1)^2 = \sqrt { x - y_2 } : \sqrt { x - y_1}\)
\(\frac {d_2}{d_1} = \sqrt[4] { \frac {x - y_2} {x - y_1} }\)
\(\frac {d_2}{d_1} = \sqrt[4] { \frac {36.9 - 17.9} {36.9 - 8.85} }\)
\(\frac {d_2}{d_1} \approx 0.907\)
\(d_2 \approx 0.907 \times d_1\)
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Grade: High School
Subject: Physics
Chapter: Kinematics
