A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill is α = 30°, and the angle of the barrel to the horizontal ß = 60°. The initial velocity v of the shell is 21 m/sec. Then distance of point from the gun at which shell will fall

The correct option is 30 m.

To find the distance from the gun to the point where the shell lands on the hill (the range along the inclined plane), we can analyze the motion of the projectile relative to the incline.

First, let us identify the given parameters:
- Slope of the hill, $\alpha ={30}^{\circ }$
- Angle of the barrel with the horizontal, $\beta ={60}^{\circ }$
- Initial velocity of the shell, $v=21\text{m/s}$
- Acceleration due to gravity, $g=9.8{\text{m/s}}^2$

The angle of projection with respect to the inclined plane, $\theta$, is:
$\theta =\beta -\alpha ={60}^{\circ }-{30}^{\circ }={30}^{\circ }$

The formula for the range $R$ of a projectile fired up an inclined plane is:
$R=\frac{2v^{2}sin(\beta -\alpha )cos(\beta )}{g{cos}^2(\alpha )}$

Substituting the given values into the range formula:
$R=\frac{2\cdot {21}^2\cdot sin\left({30}^{\circ }\right)���cos\left({60}^{\circ }\right)}{9.8\cdot {cos}^2\left({30}^{\circ }\right)}$

We know the trigonometric ratios:
- $sin\left({30}^{\circ }\right)=0.5$
- $cos\left({60}^{\circ }\right)=0.5$
- $cos\left({30}^{\circ }\right)=\frac{\sqrt{3}}{2}$, so ${cos}^2\left({30}^{\circ }\right)=\frac{3}{4}=0.75$

Substitute these values back into the equation:
$R=\frac{2\cdot 441\cdot 0.5\cdot 0.5}{9.8\cdot 0.75}$

Simplify the terms:
- Numerator: $2\cdot 441\cdot 0.25=220.5$
- Denominator: $9.8\cdot 0.75=7.35$

Now, compute the range:

Therefore, the distance from the gun to the point where the shell falls is 30 m.