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The initial kinetic energy imparted to a 0.07 kg bullet is 2341 J. The acceleration due to gravity is 9.81 m/s². Neglecting air resistance, find the range of this projectile when it is fired.

A. 142.5 m
B. 185.7 m
C. 97.3 m
D. 210.8 m

Answer :

Final answer:

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.

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.

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