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

• A

  $2 \sqrt{20} \mathrm{~m} / \mathrm{s}, 60^{\circ}$
• Correct answer
  B

  $20 \sqrt{3} \mathrm{~m} / \mathrm{s}, 60^{\circ}$
• C

  $6 \sqrt{40} \mathrm{~m} / \mathrm{s}, 30^{\circ}$
• D

  $40 \sqrt{6} \mathrm{~m} / \mathrm{s}, 30^{\circ}$

Solution

Let in 2 s body reaches upto point $A$ and after one more sec upto point $B$. Total time of ascent for a body is given 3 s
$\begin{array}{lrl} \text { i.e., } & & t \\ & =\frac{u \sin \theta}{g}=3 \\ \therefore & u \sin \theta & =10 \times 3 \\ & u \sin \theta & =30...(i) \end{array}$
Horizontal component of velocity remains always constant $u \cos \theta=v \cos 30^{\circ}$...(ii)
For vertical upward motion between point $O$ and $A$ $v \sin 30^{\circ}=u \sin \theta-g \times 2 \quad[\text { Using } v=u-g t]$ $\begin{aligned} & v \sin 30^{\circ}=30-20 \quad[A s u \sin \theta=30] \\ & \therefore \quad v=20 \mathrm{~m} / \mathrm{s} \end{aligned}$
Substituting this value $v$, we get in Eq. (ii) $u \cos \theta=20 \cos 30^{\circ}=10 \sqrt{3}$...(iii)
From Eqs. (i) and (iii) $u=20 \sqrt{3} \text { and } \theta=60^{\circ}$

Hence, the correct answer is option: 'B'.