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 horizontal distance between two bodies, when their velocity are perpendicular to each other, is.

Options

A. 1 m
B. 0.5 m
C. 2 m
D. 4 m

Answer

D

Explanation

(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`.
.