91Ó°ÊÓ

The spring constant of the spring is, k = 700 N/m

The angle of inclination is, $\mathrm{θ}=30°$

The mass of the ball is,$\mathrm{m}=57\mathrm{g}\left(\frac{{10}^{-3}\mathrm{kg}}{1\mathrm{g}}\right)=5.7×{10}^{-2}\mathrm{kg}$

The maximum height of the muzzle is, h = 1.83 m

Initial velocity can be calculated by using a kinematic equation for motion under gravity. The spring’s compression can be calculated by using the law of conservation of energy.

Formulae:

The third equation of the kinematic motion,${\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2-2\mathrm{gy}$ (1)

Applying the law of conservation of energy,

Initial total energy = Final total energy (2)

Using equation (1) for the case of launching the ball, we can get the vertical speed of the ball as given: $\left({\mathrm{v}}_{\mathrm{f}}=0\mathrm{m}/\mathrm{s}\right)$

$$\begin{gathered} 0\mathrm{m}/\mathrm{s}={\mathrm{v}}_{0\mathrm{y}}^2-\left(2×9.8\mathrm{m}/{\mathrm{s}}^2×1.83\mathrm{m}\right) \\ {\mathrm{v}}_{0\mathrm{y}}^2=35.87{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {\mathrm{v}}_{0\mathrm{y}}^2=6\mathrm{m}/\mathrm{s} \end{gathered}$$

We have,

The vertical component of initial velocity will give the launching speed that is given as:

$$\begin{gathered} {\mathrm{v}}_{0\mathrm{y}}={\mathrm{v}}_0\sin \left(\mathrm{θ}\right) \\ 6\mathrm{m}/\mathrm{s}={\mathrm{v}}_0\sin \left(30°\right) \\ {\mathrm{v}}_0=12\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the launching speed of the ball is 12 m/s .

Using equation (2), we can get the compressed distance as given:

$$\begin{gathered} \frac{1}{2}{\mathrm{kd}}^2=\frac{1}{2}{\mathrm{mv}}_0^2+\mathrm{mgdsin}\left(\mathrm{θ}\right) \\ \frac{1}{2}×700\mathrm{N}/\mathrm{m}×\left({\mathrm{d}}^2\right)=\frac{1}{2}×\left(5.7×{10}^{-2}\mathrm{kg}\right)×{\left(12\mathrm{m}/\mathrm{s}\right)}^2+\left(5.7×{10}^{-2}\mathrm{kg}\right)×9.8\mathrm{m}/{\mathrm{s}}^2×\mathrm{d}×\sin \left(30°\right) \\ \left(350\mathrm{N}/\mathrm{m}\right){\mathrm{d}}^2=4.104\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2+\left(0.2793\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{d} \\ \left(350\mathrm{N}/\mathrm{m}\right){\mathrm{d}}^2-\left(0.2793\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\right)\mathrm{d}-4.104\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2=0 \end{gathered}$$

By solving the above quadratic equation for d: we get,

d = 0.11 m

Hence, the value of the distance of compression is 0.11 m .