If t₁ and t₂ are the times of flight of two particles having the same initial velocity u and range R

Q: Wait, why are the angles complementary? Is that a rule or just a coincidence?

It is a beautiful geometric necessity! The range formula is $R = \frac{u^2 \sin(2\theta)}{g}$. For a fixed $R$ and $u$, $\sin(2\theta)$ must be constant. Since $\sin(2\theta) = \sin(180^\circ - 2\theta)$, we have $2\theta_2 = 180^\circ - 2\theta_1$, which simplifies to $\theta_1 + \theta_2 = 90^\circ$. It is not a coincidence; it is the symmetry of the sine function!

Q: Can I solve this using energy conservation instead of the time of flight formula?

Energy conservation is a powerful tool, but it tells us about speeds at specific heights, not the time spent in the air. Time of flight is fundamentally about the vertical component of motion and the acceleration due to gravity. Stick to the kinematic equations for time; they are the direct path to the solution.

Q: Does the final answer $4u^2/g^2$ mean the time of flight is independent of the angle?

Not exactly! The sum of the squares of the times is independent of the angle, but the individual times $t_1$ and $t_2$ are very much dependent on $\alpha$. The beauty here is that the 'angle-dependence' cancels out perfectly when we sum the squares. It is a classic example of how complex variables can simplify into a clean, elegant constant.

Question

If $t_1$ and $t_2$ are the times of flight of two particles having the same initial velocity u and range R on the horizontal, then $t_1^2 + t_2^2$ is equal to

Options:

(A) 1

(B) $4u^2/g^2$

(C) $u^2/2g$

(D) $u^2/g$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat plain, holding a projectile launcher. You fire a particle at an angle $\alpha$ with an initial velocity $u$. It arcs through the sky and lands at a distance $R$.

Now, you fire a second particle with the exact same velocity $u$, and it lands at the exact same spot $R$. This is the fundamental mystery of projectile motion, which forces us to confront the symmetry of the physical world.

For a given velocity $u$, two projectiles will only share the same range if their angles of projection are complementary. That is, if the first angle is $\alpha$, the second must be $90^\circ - \alpha$.

### The Mathematical Toolkit

To solve this, we need to quantify the time each particle spends in the air. The general formula for the time of flight $T$ is derived from the vertical component of motion:

$$T = \frac{2u \sin 	heta}{g}$$

Here, $	heta$ is the angle of projection. This formula tells us that the time in the air is directly proportional to the vertical component of the initial velocity.

For our first particle, projected at angle $\alpha$, the time of flight $t_1$ is:

$$t_1 = \frac{2u \sin \alpha}{g}$$

Now, consider the second particle. Its angle is $90^\circ - \alpha$. Substituting this into our formula, we get:

$$t_2 = \frac{2u \sin(90^\circ - \alpha)}{g}$$

Using the trigonometric identity $\sin(90^\circ - \alpha) = \cos \alpha$, the time of flight for the second particle simplifies to:

$$t_2 = \frac{2u \cos \alpha}{g}$$

### The Algebraic Dance

We are asked to find the value of $t_1^2 + t_2^2$. Let us square our expressions for $t_1$ and $t_2$.

Squaring $t_1$:

$$t_1^2 = \left(\frac{2u \sin \alpha}{g}ight)^2 = \frac{4u^2 \sin^2 \alpha}{g^2}$$

Squaring $t_2$:

$$t_2^2 = \left(\frac{2u \cos \alpha}{g}ight)^2 = \frac{4u^2 \cos^2 \alpha}{g^2}$$

Now, we perform the summation:

$$t_1^2 + t_2^2 = \frac{4u^2 \sin^2 \alpha}{g^2} + \frac{4u^2 \cos^2 \alpha}{g^2}$$

Notice the common factor of $\frac{4u^2}{g^2}$ in both terms. Factoring it out, we obtain:

$$t_1^2 + t_2^2 = \frac{4u^2}{g^2} (\sin^2 \alpha + \cos^2 \alpha)$$

### The Grand Finale

We have arrived at the most beautiful identity in trigonometry: $\sin^2 \alpha + \cos^2 \alpha = 1$. The entire dependence on the angle $\alpha$ vanishes, leaving us with a result that is elegant and constant.

Substituting the identity into our equation:

$$t_1^2 + t_2^2 = \frac{4u^2}{g^2} (1)$$

The final result is:

**$$t_1^2 + t_2^2 = \frac{4u^2}{g^2}$$**

This result is a testament to the underlying order of physics. No matter what angle you choose, as long as the range remains the same, the sum of the squares of the flight times remains invariant.

Animated Solution for Mathematics - Trigonometry: If $t_{1}$ and $t_{2}$ are the times of flight of two particles having the same initial velocity u and range R on the horizontal, then $t_{1}^{2} + t_{2}^{2}$ is equal to

Select Answer:

Visualized Solution

Problem Setup: Same Range and Velocity

Condition for Equal Range

Visualizing the Trajectories

General Formula for Time of Flight

Time of Flight $t_{1}$

Time of Flight $t_{2}$

Simplifying $t_{2}$

Squaring $t_{1}$

Squaring $t_{2}$

Summing the Squares

Factoring the Expression

Applying Trigonometric Identity

Final Result

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Mathematical Toolkit

The Algebraic Dance

The Grand Finale

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Time of Flight .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1​

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