JEE PYQ: Motion in a Plane - Question ID a107e47ea3f0 (JEE Main 2025)

Welcome, aspiring engineers! Let's dissect this classic projectile motion problem. It tests your fundamental understanding of range and maximum height and your proficiency with trigonometric identities.

Core Formula & Concept:

In projectile motion, when a particle is launched with an initial velocity $u$ at an angle $\theta$ with the horizontal, its trajectory is a parabola under the influence of constant gravitational acceleration $g$. The key physical quantities we'll use here are the horizontal range ($R$) and the maximum height ($H$) attained.

The standard formulas for these quantities are:

• Horizontal Range (R): This is the total horizontal distance covered by the projectile from its launch point until it returns to the same horizontal level. $R = \frac{u^2 \sin(2\theta)}{g}$
• Maximum Height (H): This is the greatest vertical distance reached by the projectile from its launch level. $H = \frac{u^2 \sin^2\theta}{2g}$

The problem provides a relationship between these two quantities ($R = 3H$) and asks us to find a specific constant $n$ by comparing the derived range with a given expression.

Step-by-Step Derivation:

Step 1: Write down the expressions for Range and Maximum Height.

As per our core concepts:

$R = \frac{u^2 \sin(2\theta)}{g} \quad \text{(Equation 1)}$ $H = \frac{u^2 \sin^2\theta}{2g} \quad \text{(Equation 2)}$

Step 2: Apply the given condition relating Range and Maximum Height.

The problem states that the horizontal range is three times the maximum height:

$R = 3H \quad \text{(Equation 3)}$

Step 3: Substitute the formulas from Equation 1 and Equation 2 into Equation 3.

Substituting the expressions for $R$ and $H$ into $R = 3H$:

$\frac{u^2 \sin(2\theta)}{g} = 3 \left( \frac{u^2 \sin^2\theta}{2g} \right)$

Step 4: Simplify the equation to find the launch angle $\theta$.

We can cancel out the common terms $u^2$ and $g$ from both sides of the equation, assuming $u \neq 0$ and $g \neq 0$:

$\sin(2\theta) = \frac{3}{2} \sin^2\theta$

Now, we use the trigonometric identity $\sin(2\theta) = 2\sin\theta\cos\theta$:

$2\sin\theta\cos\theta = \frac{3}{2} \sin^2\theta$

Assuming the particle is actually projected (i.e., $\sin\theta \neq 0$), we can divide both sides by $\sin\theta$:

$2\cos\theta = \frac{3}{2} \sin\theta$

Rearrange the terms to find $\tan\theta$:

$\frac{\sin\theta}{\cos\theta} = \frac{2 \times 2}{3}$ $\tan\theta = \frac{4}{3}$

Step 5: Determine $\sin(2\theta)$ from the value of $\tan\theta$.

Since $\tan\theta = \frac{4}{3}$, we can visualize a right-angled triangle where the opposite side is 4 units and the adjacent side is 3 units. Using the Pythagorean theorem, the hypotenuse would be $\sqrt{4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5$ units.

From this triangle, we can find $\sin\theta$ and $\cos\theta$:

$\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{4}{5}$ $\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{3}{5}$

Now, calculate $\sin(2\theta)$ using the identity $\sin(2\theta) = 2\sin\theta\cos\theta$:

$\sin(2\theta) = 2 \left( \frac{4}{5} \right) \left( \frac{3}{5} \right)$ $\sin(2\theta) = \frac{24}{25}$

Step 6: Substitute the value of $\sin(2\theta)$ back into the Range formula.

Using Equation 1 and the calculated value of $\sin(2\theta)$:

$R = \frac{u^2}{g} \left( \frac{24}{25} \right)$ $R = \frac{24 u^2}{25 g}$

Step 7: Compare the derived range with the given expression to find $n$.

The problem states that the horizontal range is given as $\frac{n u^2}{25 g}$.

Comparing our derived expression $R = \frac{24 u^2}{25 g}$ with the given form $R = \frac{n u^2}{25 g}$, we can clearly see that:

$n = 24$

Thus, the value of $n$ is 24.

Common Traps & Exam Tip:

• Trigonometric Identity Mix-up: A common mistake is to confuse $\sin(2\theta)$ with $2\sin\theta$ or $\sin^2\theta$. Always remember $\sin(2\theta) = 2\sin\theta\cos\theta$. Similarly, ensure you distinguish $\sin^2\theta$ (which is $(\sin\theta)^2$) correctly.
• Algebraic Errors: Be careful while simplifying the equation after substituting $R$ and $H$. Dividing by $\sin\theta$ is valid only if $\sin\theta \neq 0$, which is true for a projectile with non-zero height.
• Calculation of $\sin(2\theta)$ from $\tan\theta$: After finding $\tan\theta = 4/3$, some students might try to find $\theta$ using an inverse tangent function and then calculate $\sin(2\theta)$. This is unnecessary and can lead to rounding errors. It's more accurate and faster to construct a right-angled triangle to find $\sin\theta$ and $\cos\theta$ directly, then use $\sin(2\theta) = 2\sin\theta\cos\theta$.

Exam Tip: For such problems, a systematic approach is crucial. First, write down all relevant formulas. Second, set up the relationship given in the problem. Third, use trigonometric identities to simplify and solve for the unknown angle or a related trigonometric function. Finally, substitute back to find the required quantity. Always double-check your calculations, especially trigonometric manipulations.