First slide

Third law

Question

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. The recoil velocity of the gun is (g=10m/s²)

Moderate

A

0.2m/s
B

0.4m/s
C

0.8m/s
D

0.25m/s

Solution

$Range of the horizontal projection R = v \sqrt{\frac{2 h}{g}} 400 = v \sqrt{\frac{2 Ã 500}{10}} v = 40 m / s recoil velocity of the gun V = \frac{mv}{M} = \frac{1 Ã 40}{100} = 0.4 m / s$

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