Animated Solution for Physics - Kinematics: Two particles are projected from the same point with the same speed $u$ such that they have the same range $R$, but different maximum heights $h_1$ and $h_2$. Which of the following is correct?

Two Paths, One Destination

Imagine standing on a vast, open field with a ball in your hand. You throw it with a specific speed, $u$, and it lands at a distance $R$. Now, what if I told you that you could throw the ball with the exact same speed, but at a completely different angle, and it would still land in the exact same spot?

This is one of the most beautiful symmetries in projectile motion. For any given launch speed, there are exactly two angles that will get you to the same target (as long as the target is within your maximum range). One path is a high, looping arc, and the other is a low, direct dart.

The physics principle here is elegant: for two projectiles to have the same range with the same initial speed, their angles of projection must be complementary. This means if one angle is $\theta$, the other must be $90^{\circ }-\theta$.

Let's translate this physical reality into the language of mathematics. We know the standard formula for the range of a projectile is:

Because the angles are complementary, both $\theta$ and $90^{\circ }-\theta$ will yield this exact same range $R$. However, their maximum heights will be vastly different. Let's write down the expressions for the maximum heights, $h_1$ and $h_2$, for our two particles.

For the first particle launched at angle $\theta$:

For the second particle launched at angle $90^{\circ }-\theta$:

Using the basic trigonometric identity $\sin (90^{\circ }-\theta )=\cos \theta$, we can simplify the second height:

Now, we need to find a relationship between these heights and the range $R$. A great problem-solving strategy is to let the options guide you. All the options in the question involve the product $h_1{h}_2$. So, let's multiply our two height equations together!

Multiplying the numerators and denominators, we get:

This looks a bit messy, but we have a powerful tool in our mathematical arsenal: the double angle formula. We know that $\sin 2\theta =2\sin \theta \cos \theta$. By rearranging this, we can say that $\sin \theta \cos \theta =\frac{\sin 2\theta }{2}$.

Let's substitute this into our equation. Since our terms are squared, we will square the substitution as well:

Take a step back and look closely at the expression we just derived. Does a part of it look familiar?

Notice the term $\frac{u^4{\sin }^{2}2\theta }{g^2}$. If we recall our original range formula, $R=\frac{u^2\sin 2\theta }{g}$, we can see that squaring the range gives us exactly this term!

This is the "Aha!" moment. We can substitute $R^2$ directly into our height product equation:

Finally, by simply multiplying both sides by 16, we arrive at our beautifully clean, final relationship:

This elegant equation perfectly binds the horizontal reach of the projectiles to their vertical peaks, proving that even when paths diverge, the underlying physics remains deeply connected.