Question

. A particle is projected with a certain velocity. So as to have the same horizontal
'R. If t, and t, are the times taken to reach this point in the two possible way.
1) t/t2 = 2R/g 2), 12 = 2R/g
3) 11 + 12 = 2R/g
4) 11 - 12 = 2R/g​

Answer 1

Answer:

Horizontal range is same for angle of projection θθ and 90o−θ90o−θ

∴∴ Horizontal range R=u2sin2θg...(i)R=u2sin⁡2θg...(i)

∴t1=2usinθg,t2=2usin(90o−θ)g=2ucosθg∴t1=2usin⁡θg,t2=2usin⁡(90o−θ)g=2ucos⁡θg

∴t1t2=4u2sinθcosθg2=2g(2u2sinθcosθg)∴t1t2=4u2sin⁡θcos⁡θg2=2g(2u2sin⁡θcos⁡θg)

=2g(u2sin2θg)(∴sin2θ=2sinθcosθ)=2g(u2sin⁡2θg)(∴sin⁡2θ=2sin⁡θcos⁡θ)

=2Rg=2Rg (Using (i))