A particle is projected with a velocity u at two different angles which are complementary to each other. If H1 and H2 are the heights attained by them in each case, find the range of the particle.
A
4√H1H2
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B
2√H1H2
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C
√H1H2
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D
12√H1H2
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Solution
The correct option is A4√H1H2 Let θ and (90−θ) be the two complimentary angles of projections, When angle of projection is θ: Maximum height of a projectile, H1=u2sin2θ2g Range of a projectile R=2u2sinθcosθg When angle of projection is (90−θ): Maximum height of a projectile H2=u2sin2(90−θ)2g=u2cos2θ2g Range of a projectile R=2u2sin(90−θ)cos(90−θ)g =2u2sinθcosθg ∴ Range is same for both the angles of projection R2=4u2sin2θcos2θg ⇒R2=16(u2sin2θ2g)(u2cos2θ2g) ⇒R2=16H1H2 ∴R=4√H1H2