Question 255 of 308

In a spring gun having spring constant $100N{m}^{-1}$ a small ball $B$ of mass $100g$ is put in its barrel (as shown in figure) by compressing the spring through $0.05m$. 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 $2m$ above the ground. The value of $d$ is $m$.
$\left(g=10m{s}^{-2}\right)$

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Answer: 1

$\dfrac{1}{2}k{x}^{2}=\dfrac{1}{2}m{v}^{2}$
$K{x}^{2}=m{v}^{2}$
$v=x\sqrt{\dfrac{k}{m}}=0.05\sqrt{\dfrac{100}{0.1}}=0.05\times 10\sqrt{10}$
$v=0.5\sqrt{10}$
From $h=\dfrac{1}{2}g{t}^{2}$
$t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2\times 2}{10}}=\dfrac{2}{\sqrt{10}}$
$∴d=vt=0.5\sqrt{10}\times \dfrac{2}{\sqrt{10}}=1m$