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In a spring gun having spring constant $100\, {N} / {m}$ a small ball $'B'$ of mass $100\, {g}$ is put in its barrel (as shown in figure) by compressing the spring through $0.05\, {m}$. There should be a box placed at a distance $'d'$ on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of $2\, {m}$ above the ground. The value of $d$ is $....{m} .$ $\left(g=10\, {m} / {s}^{2}\right)$
$51$
$212$
$1$
$9$
Solution
$\frac{1}{2} k x^{2}=\frac{1}{2} m v^{2}$
$k x^{2}=m v^{2}$
$V=x \sqrt{\frac{k}{m}}=0.05 \sqrt{\frac{100}{1}}=0.05 \times 10 \sqrt{10}$
$v=0.5 \sqrt{10}$
From $h=\frac{1}{2} g t^{2}$
$t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 2}{10}}=\frac{2}{\sqrt{10}}$
$\therefore d=v t=0.5 \sqrt{10} \times \frac{2}{\sqrt{10}}=1\, {m}$
