A cricket ball thrown across a field is at heights h1 and h2 from the point of projection at times t1 and t2 respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is                             [WB JEE 2014]

For vertical motion,   h1=usinθt1-12gt12      (For h1)⇒  t1=h1+12gt12usinθ           ………(i)⇒  h2=usinθt2-12gt22                               (For h2)⇒  t2=h2+12gt22usinθ          ………(ii)On dividing Eq. (i) by Eq. (ii), we gett1t2=h1+12gt12/usinθh2+12gt22/usinθ⇒  h1t2-h2t1=g2t1t22-t12t2The time of flight of the ball,T=2usinθg=2g(usinθ)=2gh1+1/2gt12t1             [From Eq. (i)]=2t1h1g+t122=h1t1×2g+t1=h1t1×t1t22-t12t2h1t2-h2t1+t1=h1t1t22-ht12t2+ht12t2-h2t13t1h1t2-h2t1=h1t22-h2t12h1t2-h2t1