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Q.

Two towers AB and CD are situated a distance d apart as shown in fig. . AB  is 20 m high and CD is 30 m high from the ground. An object of mass m is thrown from the top of AB horizontally with a velocity of 10 m/s towards CD. Simultaneously another object of mass 2 m is thrown from the top of CD at an angle of 60° to the horizontal towards AB with the same magnitude of initial velocity as that of the first object. The two objects move in the same vertical plane, collide in mid-air and stick to each other
Find the position where the objects hit the ground. (g = 9.8 m/s2)

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a

103

b

203

c

153

d

none

answer is B.

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Detailed Solution

The resultant momentum of the objects in horizontal direction just before collision,

                            = m×10-2 m×10 cos 60°=0

Therefore Velocity of combined object by conservation of momentum,

                            3m×νx=0νx=0

Now y=10 +12gt2=10 sin 600.t +12gt2  t=23s

Thus the combined object falls vertically at a distance x from tower AB, where

                           x = 10 t=10×23=203 m

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