Two identical particles are projected horizontally in oppositedirections with a speed of 5 ms−1 each from the top of a talltower as shown. Assuming g = 10 ms−2, the distance betweenthem at the moment when their velocity vectors become mutuallyperpendicular is(A) 2.5 m (B) 5 m (C) 10 m (D) 20 m

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Arun Kumar · Answer

Hello v1=5i-gtj v2=-5i-gtj v1.v2=-25+(gt)^2=0 =>gt=5 =>t=0.5sec get distance using kinematics formulae. Thanks & Regards Arun Kumar Btech, IIT Delhi Askiitians Faculty

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Two identical particles are projected horizontally in oppositedirections with a speed of 5 ms−1 each from the top of a talltower as shown. Assuming g = 10 ms−2, the distance betweenthem at the moment when their velocity vectors become mutuallyperpendicular is(A) 2.5 m (B) 5 m (C) 10 m (D) 20 m

Two identical particles are projected horizontally in oppositedirections with a speed of 5 ms−1 each from the top of a talltower as shown. Assuming g = 10 ms−2, the distance betweenthem at the moment when their velocity vectors become mutuallyperpendicular is(A) 2.5 m (B) 5 m (C) 10 m (D) 20 m

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Two identical particles are projected horizontally in oppositedirections with a speed of 5 ms−1 each from the top of a talltower as shown. Assuming g = 10 ms−2, the distance betweenthem at the moment when their velocity vectors become mutuallyperpendicular is(A) 2.5 m (B) 5 m (C) 10 m (D) 20 m

Two identical particles are projected horizontally in oppositedirections with a speed of 5 ms−1 each from the top of a talltower as shown. Assuming g = 10 ms−2, the distance betweenthem at the moment when their velocity vectors become mutuallyperpendicular is(A) 2.5 m (B) 5 m (C) 10 m (D) 20 m

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