Question

The maximum range of a rifle bullet on level ground is $6000m$, The range when it is shot at an angle of 30 degrees from the horizontal is
A: $3000\sqrt 3 $m
B: 2000 m
C: 6000 m
D: 1000 m

Answer 1

Hint: The range of any particle undergoing a projectile motion is maximum for an angle of 45 degrees. The range is also independent of mass. The range is nothing but the horizontal distance covered by an object under motion.

Complete step by step answer:
First we need to find the range of the bullet;
The bullet undergoes a projectile motion and hence its range is independent of its mass. The range is given by the formula;
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$ here, R is the range, u is the initial velocity and $\theta $ is the angle of projection.
Also for maximum range of the projectile the angle of projection should be 45 degrees.
And $\sin (2 \times 45) = \sin 90 = 1$
Hence, ${R_{\max }} = \dfrac{{{u^2}}}{g}$.
We know that in the given condition maximum horizontal displacement is 6000 meters
So ${R_{\max }} = 6000 = \dfrac{{{u^2}}}{g}$ (equation: 1)
But the next condition says that the angle of projection is 30 degrees.
So, we apply the formula and we get;
$R = \dfrac{{{u^2}\sin (2 \times 30)}}{g}$
Thus, $R = \dfrac{{{u^2}\sin 60}}{g}$ (equation: 2)
From equation: 1 and equation: 2 we get;
$R = \dfrac{{6000 \times \sqrt 3 }}{2}$
Hence $R = 3000\sqrt 3 $meter.
Thus for angle 30 degrees the range will be $R = 3000\sqrt 3 $ meters.
Therefore, option A is correct.

Note: Projectile motion is independent of mass, it only depends on the initial velocity, the angle of projection and the acceleration due to gravity. Projectile motion is like an inverted parabola, the equation of trajectory will be a quadratic equation.