Explanation:
The given data is as follows.
Velocity of bullet,
= 814.8 m/s
Observer distance from marksman, d = 24.7 m
Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.
t =
(velocity in air = 343 m/s)
= 0.072 sec
Now, before the observer hears the report the distance traveled by the bullet is as follows.

= 
= 58.66
= 59 (approx)
Thus, we can conclude that each bullet will travel a distance of 59 m.
KE = (1/2)·(mass)·(speed)²
KE = (1/2)·(50 kg)·(18 m/s)²
KE = (25 kg)·(324 m²/s²)
KE = 8,100 kg-m²/s²
KE = 8,100 Joules
For a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as
L=2.7*10^30w
<h3> What is its luminosity?</h3>
Generally, the equation for the luminosity is mathematically given as
L=4*\pi^2*b
Therefore
L=4*\pi^2*b
L=4* \pi *(2.83*10^{18})*2.7*10^{-8}
L=2.7*10^30
In conclusion, the luminosity
L=2.7*10^30w
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