2. Time of Flight Calculation

Since the vertical acceleration $a_y = -g$ is constant and independent of the electric field (which is horizontal), the time of flight $T$ is determined solely by the condition that the total vertical displacement is zero ($y=0$ at $t=T$).

Using the equation of motion for the vertical component:

$$ y = (u \sin\theta) t – \frac{1}{2} g t^2 $$

Setting $y=0$ for $t=T$ (where $T \neq 0$):

$$ 0 = (u \sin\theta) T – \frac{1}{2} g T^2 $$ $$ T = \frac{2u \sin\theta}{g} $$

3. Analysis of Horizontal Motion

The particle travels forward to $x_0$, enters the field, stops momentarily in the x-direction, accelerates backwards, and travels from $x_0$ back to $x=0$.

Let’s divide the time $T$ into three segments or simply analyze the net displacement in the x-direction.

Phase 1: Motion before entering the field ($0 \to x_0$)

The velocity is constant $u_x = u \cos\theta$. The time taken to reach the boundary $x_0$ is:

$$ t_1 = \frac{x_0}{u \cos\theta} $$

Phase 2: Motion inside the field

The particle enters the field with velocity $u \cos\theta$. It experiences a retardation $a_x = -\frac{qE}{m}$. For the particle to return to $x_0$ (exiting the field), the net displacement inside the field relative to the boundary is zero. The time spent inside the field, $\Delta t$, corresponds to the time required to change velocity from $u \cos\theta$ to $-u \cos\theta$ (due to symmetry of constant acceleration deceleration/acceleration).

$$ v_{final} = v_{initial} + a_x \Delta t $$ $$ -u \cos\theta = u \cos\theta – \left(\frac{qE}{m}\right) \Delta t $$ $$ \frac{qE}{m} \Delta t = 2u \cos\theta \implies \Delta t = \frac{2mu \cos\theta}{qE} $$

Phase 3: Return from $x_0$ to Origin

The particle exits the field with velocity $-u \cos\theta$. The time to travel from $x_0$ back to $0$ is the same as the approach time $t_1$ because the speed is identical and distance is the same.

$$ t_{return} = t_1 = \frac{x_0}{u \cos\theta} $$

4. Solving for Electric Field Magnitude

The total time of flight $T$ is the sum of the times calculated in the horizontal analysis:

$$ T = t_1 + \Delta t + t_{return} $$ $$ T = \frac{2x_0}{u \cos\theta} + \frac{2mu \cos\theta}{qE} $$

We equate this to the total time derived from vertical motion ($T = \frac{2u \sin\theta}{g}$):

$$ \frac{2u \sin\theta}{g} = \frac{2x_0}{u \cos\theta} + \frac{2mu \cos\theta}{qE} $$

Canceling the factor of 2 from all terms:

$$ \frac{u \sin\theta}{g} = \frac{x_0}{u \cos\theta} + \frac{mu \cos\theta}{qE} $$

Rearranging to solve for the electric field term:

$$ \frac{mu \cos\theta}{qE} = \frac{u \sin\theta}{g} – \frac{x_0}{u \cos\theta} $$

Taking a common denominator on the RHS:

$$ \frac{mu \cos\theta}{qE} = \frac{u^2 \sin\theta \cos\theta – g x_0}{g u \cos\theta} $$

Inverting both sides:

$$ \frac{qE}{mu \cos\theta} = \frac{g u \cos\theta}{u^2 \sin\theta \cos\theta – g x_0} $$

Solving for $E$:

$$ E = \frac{mu \cos\theta}{q} \cdot \frac{g u \cos\theta}{u^2 \sin\theta \cos\theta – g x_0} $$ $$ E = \frac{mg}{q} \left[ \frac{u^2 \cos^2\theta}{u^2 \sin\theta \cos\theta – g x_0} \right] $$