Two bodies are projected from the same point with equal speeds in such directions that they both strike the same point on a plane whose inclination is `beta`. If `alpha` the angle of projection of the first body with the horizontal , show that the ratio of their time of flights is `(sin (alpha - beta))/(cos alpha)`

Explanation

Let `alpha` be the angle of projection of the second by .
`R =(u^(2))/(g cos^(2) beta) [sin (2alpha - beta)-sin beta]`
Range of both the bodies is same therefore ,
`sin(2alpha - beta) = sin (2alpha - beta) or sin 2alpha = pi- (2alpha - beta)`
`alpha = (pi)/(2) -(alpha- beta)`
Now `T = (2 u sin (alpha - beta))/(g cos beta) and T =(2 usin (alpha - beta))/(g cose beta)`
`therefore (T)/(T) = (2 sin (alpha - beta))/(2 sin (alpha - beta)) = (sin(alpha - beta))/(sin{(pi)/(2) (alpha- beta)-beta})`
`sin =((alpha - beta))/(sin((pi)/(2)-alpha)) = (sin(alpha - beta))/(cos alpha)`