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Question

A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flights in the two cases, then the product of the two times of flights is proportional to

A
1R2
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B
R2
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C
R
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D
1R
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Solution

The correct option is C R
Time of flight of projectile
T=2u sin θg
Hence t1t2=2u sin θg×2u sin αg......(i)
Since in both case range is same, α and θ is complementary angle.
α+θ=90o
From eqn. (i)
t1t2=2u sin θg×2u sin(90oθ)g
t1t2=2u sin θ×2u cos θg2
t1t2=2g×u2 sin 2θg
t1t2=2Rg
t1t2R

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