A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t$$_1$$ and t$$_2$$ are the values of the time taken by it to hit the target in two possible ways, the product t$$_1$$t$$_2$$ is:

We consider the usual equations for the motion of a projectile launched from ground level.

For an initial speed $$u$$ and an angle of projection $$\theta$$ with the horizontal, the time of flight is given by the well-known formula

$$t \;=\; \frac{2u\sin\theta}{g},$$

because the vertical component of velocity is $$u\sin\theta$$ and the projectile returns to the same vertical level after rising for $$u\sin\theta/g$$ seconds and falling for the same duration.

The horizontal range reached in that flight is, by the standard range formula,

$$R \;=\; \frac{u^{2}\sin2\theta}{g}.$$

We are told that the shell actually lands at the given distance $$R$$ from the gun, and that two different angles of projection are possible. Let those angles be $$\theta_{1}$$ and $$\theta_{2}$$, with the corresponding times of flight $$t_{1}$$ and $$t_{2}$$.

Since both angles produce the same range, we must have

$$\sin2\theta_{1} \;=\; \sin2\theta_{2} \;=\; \frac{gR}{u^{2}}.$$

Because $$\sin(180^{\circ}-\alpha)=\sin\alpha$$, the two values of $$2\theta$$ that give the same sine add up to $$180^{\circ}$$, so

$$2\theta_{1} + 2\theta_{2} = 180^{\circ} \;\;\Longrightarrow\;\; \theta_{1} + \theta_{2} = 90^{\circ}.$$

Now their times of flight, from the first formula, are

$$t_{1} = \frac{2u\sin\theta_{1}}{g}, \quad t_{2} = \frac{2u\sin\theta_{2}}{g}.$$

We multiply these two expressions to obtain the required product:

$$ t_{1}t_{2} = \left(\frac{2u\sin\theta_{1}}{g}\right)\!\!\left(\frac{2u\sin\theta_{2}}{g}\right) = \frac{4u^{2}}{g^{2}}\,\sin\theta_{1}\sin\theta_{2}. $$

Because $$\theta_{1}+\theta_{2}=90^{\circ}$$, we may write $$\sin\theta_{2}=\cos\theta_{1}$$, giving

$$\sin\theta_{1}\sin\theta_{2} = \sin\theta_{1}\cos\theta_{1} = \tfrac12\sin2\theta_{1}.$$

But from the range equality stated earlier,

$$\sin2\theta_{1} = \frac{gR}{u^{2}}.$$

Substituting this into the product of sines gives

$$\sin\theta_{1}\sin\theta_{2} = \tfrac12 \left(\frac{gR}{u^{2}}\right) = \frac{gR}{2u^{2}}.$$

Putting this back into the expression for $$t_{1}t_{2}$$, we have

$$ t_{1}t_{2} = \frac{4u^{2}}{g^{2}} \times \frac{gR}{2u^{2}} = \frac{4}{g^{2}}\times\frac{gR}{2} = \frac{2R}{g}. $$

Thus

$$t_{1}t_{2} = \frac{2R}{g}.$$

Hence, the correct answer is Option C.