JEE PYQ: Motion in a Plane - Question ID e66a0d15fb12 (JEE Main 2022)

Core Formula & Concept:

In projectile motion, an object is launched with an initial velocity \( u \) at an angle \( \theta \) with the horizontal. The motion can be resolved into two independent components:

• Horizontal motion: Uniform motion with velocity \( u \cos \theta \) (ignoring air resistance).
• Vertical motion: Motion under constant acceleration due to gravity \( g \), with initial velocity \( u \sin \theta \).

The horizontal range \( R \) of a projectile is the horizontal distance traveled before it returns to the same vertical level. The formula for the range is:

$R = \frac{u^2 \sin 2\theta}{g}$

This formula is derived from the fact that the time of flight \( T \) is:

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

and the horizontal range is:

$R = (u \cos \theta) \cdot T = u \cos \theta \cdot \frac{2u \sin \theta}{g} = \frac{2u^2 \sin \theta \cos \theta}{g} = \frac{u^2 \sin 2\theta}{g}$

using the trigonometric identity \( \sin 2\theta = 2 \sin \theta \cos \theta \).

The range \( R \) is maximum when \( \sin 2\theta \) is maximum. Since \( \sin 2\theta \) has a maximum value of 1, this occurs when:

$2\theta = 90^\circ \Rightarrow \theta = 45^\circ$

So, the maximum range \( R_{\text{max}} \) is:

$R_{\text{max}} = \frac{u^2}{g}$

Now, the question states that a second object is projected with the same initial speed \( u \), but with a range \( R_2 = \frac{R_{\text{max}}}{2} \). We are to find the angle(s) \( \theta_2 \) at which this occurs.

Step-by-Step Derivation:

Let’s denote:

• \( R_{\text{max}} = \frac{u^2}{g} \) (at \( \theta = 45^\circ \))
• \( R_2 = \frac{R_{\text{max}}}{2} = \frac{u^2}{2g} \)

We use the range formula for the second object:

$R_2 = \frac{u^2 \sin 2\theta_2}{g}$

Substitute \( R_2 \):

$\frac{u^2}{2g} = \frac{u^2 \sin 2\theta_2}{g}$

Cancel \( \frac{u^2}{g} \) from both sides:

$\frac{1}{2} = \sin 2\theta_2$

So,

$\sin 2\theta_2 = \frac{1}{2}$

The general solutions for \( \sin x = \frac{1}{2} \) are:

$x = 30^\circ + 360^\circ n \quad \text{or} \quad x = 150^\circ + 360^\circ n, \quad n \in \mathbb{Z}$

Apply this to \( 2\theta_2 \):

$2\theta_2 = 30^\circ \quad \text{or} \quad 2\theta_2 = 150^\circ$

Divide by 2:

$\theta_2 = 15^\circ \quad \text{or} \quad \theta_2 = 75^\circ$

Both angles give the same range because \( \sin 2\theta \) is symmetric about \( 90^\circ \). That is, \( \sin 2(15^\circ) = \sin 30^\circ = \frac{1}{2} \), and \( \sin 2(75^\circ) = \sin 150^\circ = \frac{1}{2} \).

Hence, the second object must be projected at either \( 15^\circ \) or \( 75^\circ \) to achieve half the maximum range.

The answer key confirms this: 15 or 75 degrees.

Common Traps & Exam Tip:

Trap 1: Students often forget that there are two angles that give the same range for a given initial speed. The range formula \( R = \frac{u^2 \sin 2\theta}{g} \) is symmetric: angles \( \theta \) and \( 90^\circ - \theta \) yield the same range. So, for every solution \( \theta \), there is a complementary solution \( 90^\circ - \theta \).

Trap 2: Misinterpreting the question: some students assume that halving the range implies halving the angle. This is incorrect. The relationship between range and angle is nonlinear due to the sine function.

Trap 3: Forgetting to consider the maximum range condition. The question states that the first object is projected at the angle for maximum range (\( 45^\circ \)), so its range is \( \frac{u^2}{g} \). The second object has range half of this, not half of an arbitrary range.

Exam Tip: Always write down the range formula and use trigonometric identities carefully. When solving \( \sin 2\theta = k \), remember to find all possible angles within the domain \( 0^\circ \leq \theta \leq 90^\circ \).

Final Answer: The angle of projection for the second object is 15° or 75°.