question_answer 71)

                  A 100 kg gun fires a ball of 1kg horizontally from a cliff of height 500m. It falls on the ground at a distance of 400m from the bottom of the cliff. Find the recoil velocity of the gun. (Acceleration due to gravity \[=10\,m{{s}^{-2}}\])

Answer:

                  Time taken by a ball to reach ground,                 \[t=\,\sqrt{\frac{2h}{g}}\,=\,\sqrt{\frac{2\times \,500}{10}}\,=10\,s.\]                 Now, Range \[=u\times t\]                 \[\therefore \] \[\upsilon \,=\,\frac{400}{10}\,\,=\,40\,\,m{{s}^{-1}}\]                 This is the velocity of the ball.                 According to the law of conservation of momentum.                 \[m\upsilon \,+\,MV=0\] or \[|V|\,\,=\,\frac{m\upsilon }{M}=\,\frac{1\times \,40}{100}\]                 \[=0.4\text{ }m{{s}^{1}}\]