A $15 \,g$ ball is shot from a spring gun whose spring has a force constant of $600 \,N/m$. The spring is compressed by $5 \,cm$. The greatest possible horizontal range of the ball for this compression is .... $m$ ($g = 10 \,m/s^2$)
Diffcult
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(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)
${R_{max}} = \frac{{k{x^2}}}{{mg}} = \frac{{600 \times {{(5 \times {{10}^{ - 2}})}^2}}}{{15 \times {{10}^{ - 3}} \times 10}}=10\,m.$
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