For a given velocity, a projectile has the same range R for two angles of projection. If t1 and t2 are the time of flight in the two cases, then t1t2 is equal to
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a
2Rg
b
Rg
c
4Rg
d
R2g
answer is A.
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Detailed Solution
R =u2sin2θg at angles θ and 90°-θNow, t1=2usinθg and t2=2usin90°-θg=2ucosθg∴ t1t2=2gu2sin2θg=2Rg
For a given velocity, a projectile has the same range R for two angles of projection. If t1 and t2 are the time of flight in the two cases, then t1t2 is equal to