A uniform electric field, $$\overrightarrow{E} = −400\sqrt{3}\hat{y} NC^{−1}$$ is applied in a region. A charged particle of mass m carrying positive charge q is projected in this region with an initial speed of $$2\sqrt{10} \times 106 ms^{−1}$$. This particle is aimed to hit a target T, which is 5 m away from its entry point into the field as shown schematically in the figure. Take $$\frac{q}{m} = 10^{10} Ckg^{−1}$$. Then

The electric field is uniform and points in the negative $$y$$-direction: $$\overrightarrow{E}= -400\sqrt3\,\hat y\;{\rm N\,C^{-1}}$$.
For a positive charge, the electric force is $$\overrightarrow{F}=q\overrightarrow{E}$$, giving an acceleration

$$\overrightarrow a=\frac{q}{m}\overrightarrow{E}= -\frac{q}{m}\,400\sqrt3\,\hat y.$$

With $$\dfrac{q}{m}=10^{10}\;{\rm C\,kg^{-1}},$$ the magnitude of the (downward) acceleration is
$$|a|=400\sqrt3\times10^{10}\;{\rm m\,s^{-2}}\approx6.928\times10^{12}\;{\rm m\,s^{-2}}.$$

The particle enters the field with speed
$$v_0 = 2\sqrt{10}\times10^6\;{\rm m\,s^{-1}} =6.324\times10^{6}\;{\rm m\,s^{-1}}.$$

Let the projection angle with the horizontal be $$\theta$$. Components of the initial velocity are $$v_{0x}=v_0\cos\theta,\qquad v_{0y}=v_0\sin\theta.$$

The target T lies 5 m horizontally from the entry point and is on the same vertical level. If $$t$$ is the time taken to hit T, the kinematic equations are

Horizontal motion: $$x(t)=v_{0x}t=5\;\Rightarrow\;t=\dfrac{5}{v_0\cos\theta}\;-(1)$$

Vertical motion: $$y(t)=v_{0y}t+\tfrac12 a t^2=0\;-(2)$$ (subscript 0 denotes initial quantities).

Substituting $$t$$ from $$(1)$$ into $$(2)$$:

$$v_0\sin\theta\;\frac{5}{v_0\cos\theta} +\frac12 a\!\left(\frac{5}{v_0\cos\theta}\right)^{\!2}=0.$$

After cancelling $$v_0$$ and simplifying,

$$5\tan\theta+\frac{25a}{2v_0^{2}\cos^{2}\theta}=0 \;\;\Longrightarrow\;\; \tan\theta=-\frac{5a}{2v_0^{2}\cos^{2}\theta}.$$

Because $$a$$ is negative (downward), write $$a=-|a|$$. Then

$$\tan\theta=\frac{5|a|}{2v_0^{2}\cos^{2}\theta}.$$

Multiply by $$\cos^{2}\theta$$:

$$\sin\theta\cos\theta=\frac{5|a|}{2v_0^{2}}.$$

Using $$\sin\theta\cos\theta=\tfrac12\sin2\theta$$ gives

$$\tfrac12\sin2\theta=\frac{5|a|}{2v_0^{2}} \quad\Longrightarrow\quad \sin2\theta=\frac{5|a|}{v_0^{2}}.$$

Compute the right-hand side:

$$\frac{5|a|}{v_0^{2}} =\frac{5(6.928\times10^{12})}{(6.324\times10^{6})^{2}} =\frac{3.464\times10^{13}}{4.000\times10^{13}} \approx0.866=\frac{\sqrt3}{2}.$$

Hence $$\sin2\theta= \frac{\sqrt3}{2}\;,$$ so

$$2\theta = 60^{\circ}\;\;{\rm or}\;\;120^{\circ} \quad\Longrightarrow\quad \theta = 30^{\circ}\;\;{\rm or}\;\;60^{\circ}.$$

Thus the particle can hit T when projected at either $$30^{\circ}$$ or $$60^{\circ}$$.

Next, compute the time of flight for each angle using $$(1)$$.

$$\cos30^{\circ}=0.866,$$ so $$t_1=\frac{5}{6.324\times10^{6}\times0.866} =9.14\times10^{-7}\;{\rm s} =0.9136\;\mu{\rm s}.$$

Numerically, $$\sqrt{\frac{5}{6}}\;\mu{\rm s} =\sqrt{0.8333}\;\mu{\rm s} =0.9129\;\mu{\rm s},$$ which matches $$t_1$$.

$$\cos60^{\circ}=0.5,$$ so $$t_2=\frac{5}{6.324\times10^{6}\times0.5} =1.582\times10^{-6}\;{\rm s} =1.582\;\mu{\rm s}.$$

Numerically, $$\sqrt{\frac{5}{2}}\;\mu{\rm s} =\sqrt{2.5}\;\mu{\rm s} =1.582\;\mu{\rm s},$$ which matches $$t_2$$.

Therefore two different times of flight are possible: $$t=\sqrt{\frac{5}{6}}\;\mu{\rm s}\quad\text{or}\quad t=\sqrt{\frac{5}{2}}\;\mu{\rm s}.$$

Checking the given statements:

Option A: angle $$45^{\circ}$$ ⇒ not satisfied. (False)

Option B: angles $$30^{\circ}$$ or $$60^{\circ}$$ ⇒ satisfied. (True)

Option C: time could be $$\sqrt{5/6}\ \mu{\rm s}$$ as well as $$\sqrt{5/2}\ \mu{\rm s}$$ ⇒ satisfied. (True)

Option D: time is $$\sqrt{5/3}\ \mu{\rm s}$$ only ⇒ not satisfied. (False)

Hence the correct options are:
Option B (projection angles) and Option C (possible times).