Answer
$d=6.87mm$
Work Step by Step
We can determine the required distance as follows:
We know that:
$m_bv_{bx}=(m_b+m_B)v_x$
$\implies (0.01)(300cos30)=(10+0)v$
This simplifies to:
$v=0.2595m/s$
Now, according to the conservation of energy
$T_1+V_1=T_2+V_2$
$\implies 0+\frac{1}{2}(m_b+m_B)v^2=0+(m_b+m_B)gh$
We plug in the known values to obtain:
$0+\frac{1}{2}(10+0.01)(0.225)^2=(0.01+10)(9.81)h$
This simplifies to:
$h=3.43mm$
Now $d=\frac{h}{sin30}=\frac{3.43}{sin30}=6.87mm$
