JEE PYQ: Motion in a Plane - Question ID e7b0ecec3333 (JEE Main 2019)

In projectile motion near Earth’s surface, the horizontal and vertical motions are independent. The trajectory can be written in two equivalent forms:

By comparing the given trajectory $y = 2x - 9x^2$ with the standard Cartesian form, we can extract $\tan\theta_0$ and $v_0$.

Step 1: Match the linear term.

The given trajectory is $y = 2x - 9x^2.$ The standard form is $y = x \tan\theta_0 - \frac{g x^2}{2 v_0^2}(1 + \tan^2\theta_0).$ Equate the coefficients of $x$: \tan\theta_0 = 2. \tag{1}

Step 2: Match the quadratic term.

Equate the coefficients of $x^2$: $-9 = -\frac{g}{2 v_0^2}(1 + \tan^2\theta_0).$ Substitute $g = 10$ and $\tan\theta_0 = 2$: $$-9 = -\frac{10}{2 v_0^2}(1 + 4) \quad\Longrightarrow\quad 9 = \frac{50}{2 v_0^2} \quad\Longrightarrow\quad v_0^2 = \frac{50}{18} = \frac{25}{9} \quad\Longrightarrow\quad v_0 = \frac{5}{3}\ \text{ms}^{-1}. \tag{2}$$

Step 3: Determine the launch angle.

From $\tan\theta_0 = 2$, we have $\sin\theta_0 = \frac{2}{\sqrt{1+4}} = \frac{2}{\sqrt{5}}, \quad \cos\theta_0 = \frac{1}{\sqrt{5}}.$ Hence $\theta_0 = \cos^{-1}\!\bigl(\tfrac{1}{\sqrt{5}}\bigr) \quad\text{or equivalently}\quad \theta_0 = \sin^{-1}\!\bigl(\tfrac{2}{\sqrt{5}}\bigr).$

Step 4: Compare with the options.

We found $\theta_0 = \cos^{-1}\!\bigl(\tfrac{1}{\sqrt{5}}\bigr), \quad v_0 = \tfrac{5}{3}\ \text{ms}^{-1}.$ This matches Option A.

1. Sign confusion in the quadratic term: Students sometimes forget the negative sign in front of the $x^2$ term, leading to incorrect values of $v_0$. 2. Angle identification: The question gives both $\cos^{-1}$ and $\sin^{-1}$ forms. One must check which inverse trigonometric function matches the derived value. 3. Unit consistency: Always ensure $g$ is in the same units as the rest of the problem (here $g = 10$ m/s²).

Exam Tip: When the trajectory is given as $y = ax + bx^2$, immediately compare it to the standard form to extract $\tan\theta_0$ and $v_0$. This avoids unnecessary parametric steps.