Answer
$1.15\space i+1.33\space j$
Work Step by Step
Please see the attached image first.
Here we use the principle of conservation of momentum.
$\vec P= Constant$
We can write,
$m_{Li}\vec V_{Li}= m_{P}\vec V_{P}+m_{\alpha}\vec V_{\alpha}$
Let's choose the x-axis along the horizontal direction & y - axis along the vertical direction. Then the two components of the momentum conservation equation become,
x component : $m_{Lix}V_{Li}= m_{P}V_{px}+m_{\alpha}V_{\alpha x}$
y component : $m_{Li}V_{Liy}= m_{P}V_{py}+m_{\alpha}V_{\alpha y}$
$\rightarrow V_{Lix}= \frac{m_{P}V_{px}+m_{\alpha}V_{\alpha x}}{m_{Li}}= \frac{m_{P}V_{P}sin\theta_{1}-m_{\alpha}V_{\alpha}sin\theta_{2}}{m_{Li}} $
Let's plug known values into this equation.
$V_{Lix}=\frac{U\times1.78\space Mm/s\times sin(24.7^{\circ}) +4U\times2.4\space Mm/s\times sin(31.5^{\circ})}{5U}$
$V_{Lix}=1.15\space Mm/s$
$\uparrow V_{Liy}= \frac{m_{P}V_{py}+m_{\alpha}V_{\alpha y}}{m_{Li}}= \frac{m_{P}V_{P}cos\theta_{1}-m_{\alpha}V_{\alpha}cos\theta_{2}}{m_{Li}}$
$V_{Ly}=\frac{4U(2.43\space Mm/s)cos(31.5^{\circ})-U(1.78\space Mm/s)cos(31.5^{\circ})}{5U}=1.33\space Mm/s$
$V_{Li}=V_{Lix}i+V_{Ly}j= 1.15\space i+1.33\space j$
