Question

1.2 great balls of fire (problem 3.1.29)
on february 15, 2013, a superbolide meteor (brighter than the sun) entered earths atmosphere over chelyabinsk, russia, and exploded at an altitude of 23.5 km. eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. the blast wave took approximately 2 minutes 30 seconds to reach ground - level. the blast wave traveled at 10° above the horizon.
(a) what was the average velocity of the blast wave?
(b) compare this with the speed of sound, which is 343 m/s at sea level.

Explanation:

Step1: Convert time to seconds

We know 1 minute = 60 seconds. So 2 minutes 30 seconds = 2×60 + 30=150 seconds.

Step2: Convert altitude to meters

The altitude is 23.5 km. Since 1 km = 1000 m, 23.5 km = 23.5×1000 = 23500 m.

Step3: Calculate the vertical - component of the distance traveled

The blast - wave traveled at an angle of 10° above the horizon. If the vertical distance (altitude) is \(y = 23500\) m, and we know that \(\sin\theta=\frac{y}{d}\), where \(d\) is the distance traveled by the blast - wave and \(\theta = 10^{\circ}\). So \(d=\frac{y}{\sin\theta}=\frac{23500}{\sin(10^{\circ})}\approx\frac{23500}{0.1736}\approx135368.66\) m.

Step4: Calculate the average velocity

The average velocity \(v=\frac{d}{t}\), where \(d\) is the distance traveled and \(t\) is the time taken. Substituting \(d\approx135368.66\) m and \(t = 150\) s, we get \(v=\frac{135368.66}{150}\approx902.46\) m/s.

Step5: Compare with the speed of sound

The speed of sound \(v_s=343\) m/s. To compare, we find the ratio \(\frac{v}{v_s}=\frac{902.46}{343}\approx2.63\).

Answer:

(a) The average velocity of the blast wave is approximately 902.46 m/s.
(b) The speed of the blast wave is approximately 2.63 times the speed of sound at sea - level.