A short spring with a spring constant of $1000N{{m}^{-1}}$ is compressed by $0.1m$. How high above the starting point will a $0.2kg$ mass rise if fired vertically by this spring?

Hint: A spring is compressed or displaced from its mean position due to which it gains some potential energy. When the spring is released and a mass is kept at the starting point, the potential energy of the spring is transferred to the mass. Therefore, equating the potential energies of both the spring and the mass, height can be calculated.
Formula used:
$\dfrac{1}{2}k{{x}^{2}}=mgh$

Complete answer:
When the spring is compressed or expanded, it is displaced from its mean position due to which it gains potential energy by virtue of its compression or expansion. When an object is kept at the starting point and the spring is released, the potential energy of the spring gets transferred to the object
The spring’s potential energy is calculated as-
$P=\dfrac{1}{2}k{{x}^{2}}$
Here, $P$ is the potential energy
$k$ is the spring constant
$x$ is the displacement of the spring
Therefore,
$\dfrac{1}{2}k{{x}^{2}}=mgh$ -(1)
Here, $m$ is the mass of the body
$g$ is the acceleration due to gravity
$h$ is the height
Given, $k=1000N{{m}^{-1}}$, $x=0.1m$, $M=0.2kg$, $g=10m{{s}^{-2}}$.
Substituting given values in eq (1) equation, we get,
$\begin{align}
  & \dfrac{1}{2}\times 1000{{({{10}^{-1}})}^{2}}=0.2\times 10h \\
 & \Rightarrow 500\times {{10}^{-2}}=2h \\
 & \therefore 2.5m=h \\
\end{align}$
The object will move to a maximum height of $2.5m$.
Therefore, the mass will rise to a height of $2.5m$ when kept at the starting point of the compressed spring.

Note:
The oscillation of spring is a harmonic motion which works on the principle that the force is directly proportional to the negative of displacement. A recoil force is developed in the spring when displaced from its mean position which brings it back to its original position. This force acts on the mass and fires it upto a certain height.