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Q.

A shell is projected from a level ground with a velocity of 20m/s at 45° to the horizontal. When the shell is at the highest point it breaks into two equal fragments. One of the fragments whose initial velocity after the explosion is zero falls vertically downward. At what distance from the point of projection does the other fragment fall? (g = 10 ms^-2)

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a

30 m

b

60 m

c

90 m

d

40 m

answer is B.

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Detailed Solution

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$R = \frac{u^{2} \sin 2 θ}{g} = 40 m \Rightarrow X_{CM} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$

Since there is no external force, the trajectory of the centre of mass should not be altered.

$R = \frac{m (R / 2) + {mx}_{2}}{2 m} \Rightarrow x_{2} = \frac{3 R}{2} = 60 m$

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A shell is projected from a level ground with a velocity of 20m/s at 45° to the horizontal. When the shell is at the highest point it breaks into two equal fragments. One of the fragments whose initial velocity after the explosion is zero falls vertically downward. At what distance from the point of projection does the other fragment fall? (g = 10 ms-2)