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Question

Two bodies are projected from the top of a tower in opposite directions with velocities of u1 and u2 simultaneously. Find the time interval when their
i) velocities are mutually perpendicular
ii) displacements are mutually perpendicular.

A
(i) t=u1u2g; (ii) t=2gu1u2
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B
(i) t=2u1u2g; (ii) t=3u1u2
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C
(i) t=u1u22g; (ii) t=43u1u2
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D
(i) t=2u1u23g; (ii) t=u1u2
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Solution

The correct option is A (i) t=u1u2g; (ii) t=2gu1u2

(i) v1=u1^igt^j
v2=u2^igt^j
When velocities are perpendicular, v1.v2=0
u1u2+g2t2=0
t=u1u2g
(ii) S1=u1t^i+12gt2^j
S2=u2t^i+12gt2^j
When displacements are perpendicular i.e. S1S2, S1.S2=0
u1u2t2+14g2t4=0
t2=4u1u2g2t=2gu1u2

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