Question

A bomb at rest at the summit of a cliff breaks into two equal fragments. One of the fragments attains a horizontal velocity of $20\sqrt{3}m/s$. The horizontal distance between the two fragments, when their displacement vectors is inclined at ${60}^0$ relative to each other is (g=10\,m/s^2$

• A. $40\sqrt{3}m$
• B. $80\sqrt{3}m$
• C. $120\sqrt{3}m$
• D. $480\sqrt{3}m$

Answer 1

The correct option is D $480\sqrt{3}m$

$\left(20\sqrt{3}\right)t=d$

$S=0+\frac{1}{2}\times g{t}^2$

$S=5t^2$

Now, $tan{30}^0=\frac{d}{s}=\frac{20\sqrt{3}t}{5t^2}$

$\frac{1}{\sqrt{3}}=\frac{4\sqrt{3}}{t}$

$t=12s$

$d=20\sqrt{3}\times 12$

So, $2d=480\sqrt{3}m$