Question

From a tower of height `40 m`, two bodies are simultaneously projected horizontally in opposite direction, with velocities `2 m s^-1 and 8 ms^-1`. respectively.
The time taken for the displacement vectors of two bodies to be come perpendicular to each other is.
A. 0.1 s
B. 0.2 s
C. 0.8 s
D. 0.6 s

Answer 1

Correct Answer - C
(d) Velocity of first body at any instant t, `vec v_1 = 2 hat i- "gt" hat j`
Velocity of second body at any instant
`vec v_2 = -8 hat i- "gt" hat j`
Since `vec v_1 _|_ vec v_2, i.e., vec v_1 . vec v_2 = 0`
`(2 hat i- "gt" hat j).(-8 hat i- "gt" hat j) = 0 rArr -16 + g^2 t^2 = 0`
`rArr t = sqrt((16)/(g^2)) rArr t = (4)/(10) = 0.4 s`
`vec S_1 = (2 xx 0.4) hat i, vec S_2 =(8 xx 0.4)(- hat i)`
`vec S_1 = 0.8 hat i and vec S_2 = -3.2 hat i`
Separation `= vec S_1 - (-vec S_2) = 0.8 hat i+ (3.2 hat i) = 4 hat i`
`vec S_1 = 2t hat i -(1)/(2) "gt"^2 hat j, vec S_2 = -8t hat i - (1)/(2) "gt"^2 hat j`
As `vec S_1 _|_ vec S_2, i.e., vec S_1 . vec S_2 = 0`
`-16t + (1)/(2) xx (1)/(2) g^2 t^2 = 0 rArr g^2 t^2 = 16 xx 4`
`rArr gt = 4 xx 2 rArr t = (8)/(10) = 0.8 s`.
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