Two objects are located at height 10 m above the ground. At some point of time, the objects are thrown with initial velocity `2sqrt(2)` m/s at an angle `45^(@)` and `135^(@)` with the positive X-axis, respectively. Assuming `g = 10 m/s^(2)`, the velocity vectors will be perpendicular to each other at time is equal to  

Options

A. 0.2 s
B. 0.4 s
C. 0.6 s
D. 0.8 s

Answer

B

Explanation

Given , `U_1 = U_2 = 2 sqrt2 m//s, g= 10m//s^2`.
The velocity vectors are perpendicular to each other at time t.
`overset(to)(v_1) ( 2 sqrt(2) ((1)/(sqrt2)))overset(^)(i) 6 ( 2 sqrt(2) ((1)/(sqrt2))- 10t ) overset(^)(j)`
`= 2 overset(^)(i) + (2 - 10t)overset(^)(j)" "...(1)`
`overset(to)(v_2) =(U_2 cos 135^@ ) overset(^)(i) + (U_2 sin 135^@ - 10t ) overset(^)(j)`
`= - 2overset(^)(i) + (2 - 10t)overset(^)(j)`
`=-2 overset(^)(i) + (2 - 10t) overset(^)(j)" "...(2)`
`overset(to)(v_1). overset(to)(v_2) =0 ( therefore " Both are perpendicular")`
`rArr [2 overset(^)(j) + (2 - 10t) overset(^)(j) ] . [-2 overset(^)(i) + (2 - 10t)overset(^)(j)]=0`
`rArr -4 (2 - 10t)^2 =0`
On solving,
`100t^2 - 40t =0`
`10t (10t - 4) =0`
`therefore t=0, 0.4 sec.`