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.
(acceleration due to gravity = 10 ms-2ms−2)
Mass of the ball (m) = 1 kg
Height of the cliff (h) = 500 m
Horizontal distance traveled by the ball (x) = 400 m
From h = 12gt2(∵ Initial velocity in downward direction is zero) 500 = 12×10t2 t = √100 = 10sStep 2: Find horizontal speed of ballFrom x = ut, u = xt = 40010 = 40 m/s.From h = 12gt2(∵ Initial velocity in downward direction is zero) 500 = 12×10t2 t = √100 = 10sStep 2: Find horizontal speed of ballFrom x = ut, u = xt = 40010 = 40 m/s.
Step 2: Find the recoil velocity of the gun by applying momentum conservation.
If v is the recoil velocity of the gun, then according to the principle of conservation of linear momentum,
m1v=m2um1v=m2u
v = m2um1=1100×40=0.4m/sv = m2um1=1100×40=0.4m/s
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