Question

Two particles $A$ and $B$ are placed as shown in Fig. A.12. The particle $A$, on the top of tower, is projected horizontally with a velocity $u$ and particle $B$ is projected along the surface towards the tower, simultaneously. If both particles meet each other, then the speed of projection of particle $B$ is [ignore any friction]

• A. $d\sqrt{\frac{g}{2H}}u$
• B. $d\sqrt{\frac{g}{2H}}$
• C. $d\sqrt{\frac{g}{2H}}-u$
• D. $u$

Answer 1

The correct option is C $d\sqrt{\frac{g}{2H}}-u$

Since the two particles meet, so let's solve the problem in the non-inertial frame of B.

the velocity of A w.r.t. B $=v+u$

So, writing the equation of Motion in the X-Y direction

Along X-axis,

$(u+v)t=d$

$⟹t^2=d^2/(u+v{)}^2$

Along Y-axis

$H=1/2g{t}^2$

$⟹2H/g=d^2/(u+v{)}^2$

$⟹v=d\sqrt{\frac{g}{2H}}-u$