It is possible to project a particle with a given speed in two possible ways so as to have the same range, R. The product of the times taken to reach this point in the two possible ways is proportional to
R
The range of a projectile is given by R=2v20 cos θ sin θg.
Now, the range at complementary angles θ and 90∘−θ is equal. The times of flight coresponding to these two angles are
t1=2v0 sin θg........(1)
and t2=2v0 sin(90∘−θ)g=2v0 cos θg........(2)
From equations (1) and (2), t1t2=4v20 sin θ cos θg2=2Rg
Thus, t1t2∝R and the correct choice is (b).