Let's call the projection angle we're trying to find θ.
In the first scenario, the bullet is shot 4km away at an angle of 30°. We can use the range formula to find the muzzle speed of the bullet:
range = (muzzle speed)^2 * sin(2θ) / g,
where g is the acceleration due to gravity.
In this case, the range is 4km (4000 m), and the angle θ is 30°. Assuming the acceleration due to gravity is 9.8 m/s², let's solve for the muzzle speed:
4000 = (muzzle speed)^2 * sin(2 * 30°) / 9.8
4000 * 9.8 = (muzzle speed)^2 * sin(60°)
39200 / sin(60°) = (muzzle speed)^2
(39200 / sin(60°)) = (muzzle speed)^2
muzzle speed = sqrt(39200 / sin(60°))
Now, in the second scenario, we want to hit a target 6km (6000 m) away. To find the projection angle θ, we can rearrange the range formula:
range = (muzzle speed)^2 * sin(2θ) / g
Solving for sin(2θ):
sin(2θ) = range * g / (muzzle speed)^2
Substituting the given values:
sin(2θ) = 6000 * 9.8 / sqrt(39200 / sin(60°))^2
sin(2θ) = 6000 * 9.8 / (39200 / sin(60°))
sin(2θ) = 6000 * 9.8 * sin(60°) / 39200
sin(2θ) = 0.866 / 6.28
sin(2θ) ≈ 0.1379
Next, we can find 2θ by taking the inverse sine of 0.1379:
2θ ≈ sin^(-1)(0.1379)
θ ≈ sin^(-1)(0.1379) / 2
Calculating the projection angle using a calculator, we find:
θ ≈ 7.92°
Therefore, the projection angle needed to hit the target 6km away is approximately 7.92°.
A bullet shot hits the ground 4km away, which was at an angle of 30° with the plane. What should be the projection angle to hit a target 6km away ? Neglect air resistance and assume it's muzzle speed
1 answer