A cricket ball thrown across a field is at heights h_1 and h_2 from the point of projection at times t_1 and t_2 respectively after the throw. The ball is caught by a fielderat the same height as that of projection. Time of flight of the ball in this journey is

A cricket ball thrown across a field is at heights h_1 and h_2 from th

1.	A cricket ball thrown across a field is at heights h_1 and h_2 from the point of projection at times t_1 and t_2 respectively after the throw. The ball is caught by a fielderat the same height as that of projection. Time of flight of the ball in this journey is
Answer» `(h_1t_2^2-h_2t_1^2)/(h_1t_2-h_2t_1)` `(h_1t_2^2+h_2t_2^2)/(h_2t_1+h_1t_2)` `(h_1t_2^2+h_2t_1^2)/(h_1t_2+h_2t_1)` `(h_1t_1^2-h_2t_2^2)/(h_1t_1-h_2t_2)` Solution :`h_1=(U sin theta)t_1-1/2"gt"_1^2,h_2=(usin theta)t_2-1/2"gt"_2^2` `THEREFORE (h_1+1/2"gt"_1^2)/(h_2+1/2"gt"_2^2)=(t_1)/(t_2) or, h_1t_2-h_2t_1=g/2 (t_1t_2^2-t_1t^2t_2)` So, time of flight is GIVEN by `T=(2usin theta)/(g)=2/g[(h_1+1/2"gt"_1^2)/(t_1)] =2/(t_1)[(h_1)/(g)+(t_1^2)/(2)]` `(h_1)/(t_1)xx((t_1t_2^2-t_1^2t_2)/(h_1t_2-h_2t_1))+t_1=(h_1t_2^2-h_2t_1^2)/(h_1t_2-h_2t_1)`