Denote the (positive) speeds of objects just after explosion as `V_1` and `V_2` , and their masses as `m_1` and `m_2` . The directions of `V_1` and `V_2` are the opposite.
Then conservation of impulse gives us `m_1 V_1=m_2 V_2,` because the speed just before the explosion was zero. The conservation of energy gives `m_1 V_1^2+m_2 V_2^2=J` (the chemical energy that becomes kinetic).
Solving this system of equations is easy:
`V_1=m_2/m_1 V_2,` substitute it to the second equation and obtain `m_2/m_1 (m_2+m_1) V_2^2=2 J,` so
`V_2^2=2 J m_1/m_2 1/(m_1+m_2)` and `V_1^2=2 J m_2/m_1 1/(m_1+m_2).`
In numbers, they will be equal to
`V_1=sqrt(1800/1.7 0.3/1.4) approx 15(m/s),` `V_2=sqrt(1800/1.7 1.4/0.3) approx 70(m/s).`
For the second part of the question we have to consider an angle between the velocity of the first object and a horizontal line, and write the equations of horizontal and vertical movements of both bodies.
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