Two particles are projected from the top of a tower with velocities 10m/s and 80m/s in horizontal but in opposite directions. After what time t their velocity vectors will be mutually perpendicular to each other? (Take g=10m/s2)
A
2√2sec
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B
2sec
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C
8sec
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D
4sec
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Solution
The correct option is A2√2sec
Let u1 and u2 be the initial velocities of two particles projected from the top of the tower. →u1=80^im/s and →u2=−10^im/s
Let t be the time after which their velocity vectors are mutually perpendicular. At time t, velocity of both the particles can be given by →v1=80^i−(gt)^j and →v2=−10^i−(gt)^j
For the two velocity vectors to be perpendicular, we can write →v1.→v2=0 ⟹(80^i−(gt)^j).(−10^i−(gt)^j)=0 ⟹−800+g2t2=0 ⟹t=√8=2√2sec