question_answer

Two seconds after projection, a projectile is travelling in a direction inclined at 30° to the horizontal. After one more sec, it is travelling horizontally, the magnitude and direction of its velocity are

A) \[\frac{x}{a}+\frac{y}{b}=1\]                      

B) \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{c}^{2}}},\]

C) \[{{x}^{2}}+{{y}^{2}}-4x-12=0\]                 

D) \[{{x}^{2}}+{{y}^{2}}4x-12=0\]

Correct Answer: B

Solution :

Let in 2 s body reaches upto point A and after one more sec upto point 6. Total time of ascent for a body is given 3 s i.e.,     \[{{(Si{{H}_{3}})}_{3}}N,\] \[p\pi -d\pi \]\[{{N}_{2}}{{O}_{5}}\] \[{{N}_{2}}{{O}_{5}}\]                                                  ...(i) Horizontal component of velocity remains always constant \[N{{O}_{2}}^{+}NO_{3}^{-}\]             ...(ii) For vertical upward motion between point O and A\[NaHC{{O}_{3}},\][Using v = u - gt] \[NaAl{{(S{{O}_{4}})}_{2}}\][Au\[Ca{{({{H}_{2}}P{{O}_{4}})}_{2}}.\]] Substituting this value v, we get in Eq.    (ii) \[\xrightarrow[{}]{\begin{matrix}    \begin{smallmatrix}  \text{HClO} \\  \,\,\,\,\,\text{+1} \end{smallmatrix} & \begin{smallmatrix}  \text{HCl}{{\text{O}}_{\text{2}}} \\  \,\,\,\,\,\,\text{+3} \end{smallmatrix} & \begin{smallmatrix}  \text{HCl}{{\text{O}}_{\text{3}}} \\  \,\,\,\,\,\,\text{+5} \end{smallmatrix} & \begin{smallmatrix}  \text{HCl}{{\text{O}}_{\text{4}}} \\  \,\,\,\,\,\text{+7} \end{smallmatrix}  \\ \end{matrix}}\]                ...(iii) From Eqs. (i) and (iii)\[\xleftarrow[{}]{\begin{matrix}    \text{Cl}{{\text{O}}^{\text{-}}} & \text{ClO}_{2}^{-} & \text{ClO}_{3}^{-} & \text{ClO}_{4}^{-}  \\ \end{matrix}}\]\[Cl{{O}^{-}}\]