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=2g√u1u2
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B
(i) t=√2u1u2g; (ii) t=3√u1u2
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C
(i) t=√u1u22g; (ii) t=43√u1u2
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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=2g√u1u2
(i) →v1=−u1^i−gt^j →v2=u2^i−gt^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. →S1⊥→S2,→S1.→S2=0 ⇒−u1u2t2+14g2t4=0 ⇒t2=4u1u2g2⇒t=2g√u1u2