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Home » 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 the bottom of the cliff. Find the recoil velocity of the gun.
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 the bottom of the cliff. Find the recoil velocity of the gun.
CBSE | Class 11 | Laws of Motion | ncert exemplar | Physics
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 the bottom of the cliff. Find the recoil velocity of the gun.

The ball’s horizontal speed is = u m/s.

The vertical component is equal to zero.

Considering the ball’s vertical downward motion

\begin{array}{l} u=0 \\ s=h=500 \mathrm{~m} \\ g=10 \mathrm{~s} / \mathrm{m}^{2} \\ s=u t+1 / 2 \mathrm{at}^{2} \end{array}

We obtain by substituting the values.

\begin{array}{l} t^{2}=100 \\ t=10 \mathrm{sec} \end{array}
Horizontal range =10 \mathrm{u}
Which means v=40 \mathrm{~m} / \mathrm{s}

Using conservation of momentum, we can find the gun’s recoil velocity as = -40/100 m/s = -0.4 m/s, which is the inverse of the ball’s speed.

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