Physics - NLM Question with Solution | TestHub

At $t=\sqrt{6}$ s the boy's speed is $u=at=3\sqrt{6}$ m/s.

From the string geometry (figure): the string makes ${\theta }_1=30^{\circ }$ at the boy's end (3 m vertical, 6 m hypotenuse) and ${\theta }_2=37^{\circ }$ at the block (3-4-5 triangle).

String-length constraint: $v\cos {\theta }_2=u\sin {\theta }_1$.

$v=\frac{3\sqrt{6}\times \sin 30^{\circ }}{\cos 37^{\circ }}=\frac{3\sqrt{6}\times (1/2)}{4/5}=\frac{15\sqrt{6}}{8}$ m/s - (B)