Answer :
Final answer:
The range of the projectile when fired at a 45-degree angle is approximately 6824.49 meters.
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Explanation:
To find the range of the projectile, we can use the formula for the range R of a projectile launched at an angle (\( \theta \)) with an initial velocity [tex](\( v_0 \))[/tex]:
[tex]\[ R = \frac{{v_0^2 \sin(2\theta)}}{g} \][/tex]
Given:
Mass of the bullet ( m ) = 0.07 kg
Initial kinetic energy ( KE ) = 2341 J
Acceleration due to gravity ( g ) = 9.81 m/s²
First, we need to find the initial velocity [tex](\( v_0 \))[/tex] of the bullet using the kinetic energy:
[tex]\[ KE = \frac{1}{2}mv_0^2 \][/tex]
[tex]\[ v_0^2 = \frac{2KE}{m} \][/tex]
[tex]\[ v_0^2 = \frac{2 \times 2341}{0.07} \][/tex]
[tex]\[ v_0^2 = 66942.857 \][/tex]
[tex]\[ v_0 = \sqrt{66942.857} \][/tex]
[tex]\[ v_0[/tex] ≈ 258.8 m/s
Now, to find the range, we need to determine the angle [tex](\( \theta \))[/tex] at which the projectile is launched. Since the angle is not given, we'll assume it's launched at a 45-degree angle, which maximizes the range:
[tex]\[ \theta = 45^\circ \][/tex]
Now we can plug in the values into the range formula:
[tex]\[ R = \frac{{(258.8)^2 \sin(90)}}{9.81} \][/tex]
[tex]\[ R = \frac{{66942.857 \times 1}}{9.81} \][/tex]
R ≈ 6824.49 m
So, the range of the projectile when fired at a 45-degree angle is approximately 6824.49 meters.
The answer is not in the selected option.
