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Two particles A & B are shot from the same height at t = 0 in opposite direction with horizontal velocities 3m/s and 4m/s respectively . If they are subjected to the same vertical acceleration due to gravity (9.8), the distance between them when their velocity vectors become mutually perpendicular is (a) 1.059 m (b) 1.412m (c) 2.474m (d) 9.8m

Two particles A & B are shot from the same height at t = 0 in opposite direction with horizontal velocities
3m/s and 4m/s respectively . If they are subjected to the same vertical acceleration due to gravity (9.8), the distance between them when
their velocity vectors become mutually perpendicular is
(a) 1.059 m  (b) 1.412m  (c) 2.474m  (d) 9.8m

Grade:11

2 Answers

Akshay
185 Points
8 years ago
Let velocity of A at such instance be (3 i + V j) and of B be (-4 i + V j) where V is vertical velocity which will be same for A and B as both are thrown at same instant.
(3 i + V j).(-4 i + V j) =0. So, V2 = 12. Suppose such incident happens at t=t’.
In vertical direction, V = 0 + 9.8*t, t=rt(12)/9.8.
Relative velocity of A wrt B = 7 i . Hence, distance = 7*rt(12)/9.8 = 2.474
isha tated
11 Points
7 years ago
can u pls give a detailed explanation of Tye last step as i m not getting the calculations after relative velocity. and what is r ? is it the resultant? if yes then of what is it the resultant ?

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