b
(b) For getting horizontal range, there must be some inclination of spring with ground to project ball.

\({R_{max}} = \frac{{{u^2}}}{g}\)      …..(i)

But \(K.E.\) acquired by ball \(= P.E.\) of spring gun

\( \Rightarrow \frac{1}{2}m{u^2} = \frac{1}{2}k{x^2}\)

==> \({u^2} = \frac{{k{x^2}}}{m}\)…..(ii)

From equation (i) and (ii)

\({R_{max}} = \frac{{k{x^2}}}{{mg}} = \frac{{600 \times {{(5 \times {{10}^{ - 2}})}^2}}}{{15 \times {{10}^{ - 3}} \times 10}}=10\,m.\)