Animated Solution for Physics - Kinematics: A projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flights in the two cases, then the product of the two times of flights is proportional to

Have you ever thrown a paper ball into a dustbin? Sometimes you throw it with a low, fast trajectory, and it goes right in. Other times, you lob it high up into the air, it traces a beautiful, tall parabola, and drops perfectly into the exact same bin. This isn't just a coincidence; it's a fundamental law of physics!

In the fascinating world of kinematics, a projectile can achieve the exact same horizontal range $R$ for two different angles of projection, provided the initial speed $u$ remains constant. This is a concept that has puzzled and amazed students for generations, and today, we are going to tear it apart and understand the beautiful mathematics that makes it possible.

Imagine you are standing on a vast, flat field with a water hose. If you point the hose at a $30^{\circ }$ angle, the water lands at a certain distance. If you raise the hose to a $60^{\circ }$ angle, the water shoots much higher, but surprisingly, it lands at the exact same distance!

The mathematical condition for this to happen is beautifully simple: the two angles must be complementary. This means they must add up to $90^{\circ }$. If one angle is $\theta$, the other must be $90^{\circ }-\theta$.

Why does this work? Let's look at the master formula for the horizontal range of a projectile:

If we replace $\theta$ with $90^{\circ }-\theta$, the sine term becomes $\sin (2(90^{\circ }-\theta ))$. Expanding the term inside the sine function gives us $\sin (180^{\circ }-2\theta )$.

From basic trigonometry, we know the identity $\sin (180^{\circ }-x)=\sin x$. Therefore, $\sin (180^{\circ }-2\theta )$ is simply equal to $\sin 2\theta$. The range remains exactly the same! The math perfectly mirrors the physical reality.

A Tale of Two Trajectories

While the horizontal range is identical, the time the projectile spends in the air is vastly different. The higher you throw it, the longer it stays airborne. The time of flight is governed entirely by the vertical component of the initial velocity.

Let's write down the time of flight for both cases. For the first angle $\theta$, the vertical velocity is $u\sin \theta$. The time it takes to go up and come back down is given by:

For the second angle $90^{\circ }-\theta$, the vertical velocity is $u\sin (90^{\circ }-\theta )$. We substitute this into the time of flight formula:

Here, we use another fundamental trigonometric identity: $\sin (90^{\circ }-\theta )=\cos \theta$. This simplifies our second time of flight to:

Notice how the sine and cosine functions swap roles. The low trajectory has a small time of flight because $\sin \theta$ is small for small angles. The high trajectory has a large time of flight because $\cos \theta$ is large for small angles (meaning $90^{\circ }-\theta$ is large).

The question asks us to find the relationship involving the product of these two times of flight. In physics, multiplying two seemingly unrelated quantities often reveals a hidden invariant or a deeper physical truth. So, let's multiply $t_1$ and $t_2$ together.

Multiplying the numerators and denominators, we get:

Now, we need to be observant. Does this expression remind you of anything? In physics derivations, you must always keep one eye on the final goal. Our goal is to relate this product back to the horizontal range $R$.

Let's rearrange our product slightly to reveal a hidden structure. We can factor out a $\frac{2}{g}$ from the expression:

Look closely at the term inside the parenthesis. Using the double angle identity $\sin 2\theta =2\sin \theta \cos \theta$, the term inside the bracket transforms beautifully:

This is exactly the formula for the horizontal range $R$! The math has guided us right back to where we started, but with a profound new insight.

Substituting $R$ back into our equation, we arrive at a beautifully elegant result:

Since $2$ and $g$ are constants for any given planet, we can clearly see that the product of the times of flight is directly proportional to the horizontal range:

This means that no matter what initial speed you choose, or what specific complementary angles you use, the product of the times the two projectiles spend in the air will always scale linearly with how far they travel horizontally.

The Hidden Symmetries

This is a classic JEE concept, but the rabbit hole goes deeper. What if, instead of the time of flight, we looked at the maximum heights $H_1$ and $H_2$ reached in the two cases?

The maximum height is given by $H=\frac{u^2{\sin }^2\theta }{2g}$. For the two complementary angles, the heights would be:

If you multiply them, you will find another magical relationship:

Furthermore, if you add them, you get: This sum is completely independent of the projection angle!

Physics is full of these hidden symmetries. The more you manipulate the equations, the more secrets they reveal. Keep exploring, keep questioning, and never stop enjoying the math behind the motion!