Questions
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]
detailed solution
Correct option is C
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-h2t1Talk to our academic expert!
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