A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on

Q: Why are there two angles for the same range?

Because the range formula depends on $\sin 2\theta$. Since $\sin(180^\circ - 2\theta) = \sin 2\theta$, an angle of $(90^\circ - \theta)$ will produce the same sine value as $\theta$, resulting in the same horizontal range.

Q: What if the angle is exactly 45 degrees?

If the angle is $45^\circ$, its complementary angle is also $45^\circ$. In this unique case, there is only one trajectory, and it corresponds to the maximum possible range for that initial speed.

Q: Can the maximum heights be the same for these two angles?

No. The maximum height depends on $\sin^2 \theta$. Since $\theta$ and $90^\circ - \theta$ have different sine values (unless $\theta = 45^\circ$), the two trajectories will reach different maximum heights.

Q: Is the initial speed $u$ required to find the product $t_1 t_2$?

No! As we derived, the product $t_1 t_2 = \frac{2R}{g}$. The initial speed $u$ and the specific angles cancel out or get absorbed into the variable $R$, leaving a direct relationship between the times and the range.

Q: Does this hold true if air resistance is considered?

No. This elegant mathematical symmetry relies entirely on the idealized parabolic motion where air resistance is zero. In the real world, air drag breaks this perfect complementary relationship.

Question

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

Options:

(A) $\frac{R}{4g}$

(B) $\frac{R}{g}$

(C) $\frac{R}{2g}$

(D) $\frac{2R}{g}$

Sigma Percentile · Accepted Answer

### The Magic of Complementary Angles

Imagine you are an artillery commander. You need to hit an enemy bunker located at a precise horizontal distance, $R$. You set your cannon to fire with a specific initial speed, $u$. Interestingly, physics offers you a choice: there are exactly two different angles of elevation that will land your shell right on the target!

This happens because the horizontal range of a projectile is given by the formula $R = \frac{u^2 \sin 2	heta}{g}$. Since $\sin(180^\circ - x) = \sin x$, it follows that $\sin 2	heta_1 = \sin 2(90^\circ - 	heta_1)$. Therefore, any two angles that add up to $90^\circ$ (complementary angles) will yield the exact same horizontal range for a given initial speed.

### Analyzing the Times of Flight

Let's call these two complementary angles $	heta_1$ and $	heta_2$, where $	heta_2 = 90^\circ - 	heta_1$. Even though the shells hit the same spot, they take completely different paths. One path is lower and faster, while the other is higher and takes longer. Let's write down the time of flight for each trajectory.

For the first angle, the time of flight is:
$$t_1 = \frac{2u \sin	heta_1}{g}$$

For the second angle, we substitute $	heta_2 = 90^\circ - 	heta_1$ into the time of flight formula:
$$t_2 = \frac{2u \sin(90^\circ - 	heta_1)}{g} = \frac{2u \cos	heta_1}{g}$$

### The Master Calculation

The problem asks us to find the product of these two times, $t_1 t_2$. Let's multiply our two expressions together. Don't rush the algebra; let's look for patterns.

$$t_1 t_2 = \left( \frac{2u \sin	heta_1}{g} ight) \left( \frac{2u \cos	heta_1}{g} ight)$$

Multiplying the numerators and denominators gives:
$$t_1 t_2 = \frac{4u^2 \sin	heta_1 \cos	heta_1}{g^2}$$

Now, we need to be clever. We want to relate this back to the range $R$. Let's factor out a $\frac{2}{g}$ to expose a familiar trigonometric identity inside the remaining terms:
$$t_1 t_2 = \frac{2}{g} \left( \frac{u^2 (2\sin	heta_1 \cos	heta_1)}{g} ight)$$

### The Elegant Conclusion

Notice the term in the parentheses! From trigonometry, we know that $2\sin	heta_1 \cos	heta_1 = \sin 2	heta_1$. Let's substitute this back in:

$$t_1 t_2 = \frac{2}{g} \left( \frac{u^2 \sin 2	heta_1}{g} ight)$$

The entire expression inside the bracket is exactly the formula for the horizontal range, $R$. Therefore, we can replace it entirely:

**$$t_1 t_2 = \frac{2R}{g}$$**

This is a beautiful, standard result in kinematics. It tells us that whenever two projectiles share the same range, the product of their times of flight is directly proportional to that range, completely independent of the specific angles used!

Select Answer:

Visualized Solution

\text{Visualizing the Trajectories}

\text{Condition for Same Range}

\text{Time of Flight Equations}

\text{Product of Times of Flight}

\text{Simplifying the Product}

\text{Relating to Range}

\text{Conclusion}

The Sigma Insight: Projectile Motion

The Magic of Complementary Angles

Analyzing the Times of Flight

The Master Calculation

The Elegant Conclusion

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