= Solution
Immediately after the parcel leaves, the <spring> still stores $\tfrac12kx^2$, but only the $m_s=0.300\,\mathrm{kg}$ sledge is launched. Choose <gravitational potential> zero at the ramp base. Its initial position is $-x$, so <conservation of energy> to the apex gives
$$
\frac12kx^2-m_sgx=m_sgH_s.
$$
Therefore
$$
\boxed{H_s=\frac{kx^2}{2m_sg}-x=2.17\,\mathrm m}
$$
above the ramp base. Its upward travel from the compressed position is $H_s+x=2.56882\,\mathrm m$, a different distance from the height above the base.
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