Particle is projected with initial velocity at an angle such that it grazes the top of two poles of same height at time 2 and 6 sec . First pole is at distance 35 m from the point of projection. Then match the list-I and List-II :

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Time of flight = T OA ​ + T AB ​ + T BC ​ = 2 + 4 + 2 = 8 sec height of pole is equal at t = 2& t = 6 sec h = v y ​ × 2 − 2 1 ​ g × 2 2 = v y ​ × 6 − 2 1 ​ g × 6 2 180 − 20 = 4 v y ​ 160 = 4 v y ​ v y ​ = 40 v x ​ = 2 35 ​ H = 2 g v 2 ​ = 20 1600 ​ = 80 m R = 2 35 ​ × 10 40 × 2 ​ R = 140 DE = 140 − 70 = 70 m

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