A $100\, kg$ gun fires a ball of $1\, kg$ horizontally from a cliff of height $500 \,m $. It falls on the ground at a distance of $400 \,m $ from bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity $g = 10\,ms^{-1}$ )

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Mass of gun $m_{1}=100 \mathrm{~kg}$

mass of ball $m_{2}=1 \mathrm{~kg}$

Horizontal distance travelled by the ball, $x=400 \mathrm{~m}$

Initially vertical component of velocity is zero (in downward direction)

$h=\frac{1}{2} g t^{2}$

$500=\frac{1}{2}(10) t^{2}$

$500=5 t^{2}$

$t^{2}=100$

$t=\sqrt{100}=10 \mathrm{~s}$

For horizontal motion of ball,

$x=u t$,

$u=\frac{x}{t}=\frac{400}{10}=40 \mathrm{~m} / \mathrm{s}$

For hor $x=u t$ $t=\frac{x}{t}=\frac{400}{10}=40 \mathrm{~m} / \mathrm{s}$ Let $v$ is recoil velocity of gun,

By law of conservation of linear momentum, $m_{1} v=m_{2} u$ $\therefore v=\frac{m_{2} u}{m_{1}}=\frac{1}{100} \times 40=0.4 \mathrm{~m} / \mathrm{s}$

886-s186

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