Question

A particle is thrown in upward direction with initial velocity V. It crosses point P at height h at timr t1 and t2 . Then t1 and t2 is

Answer 1

when particle is thrown upward it occupies same position while going up and down after t1, t2 d = V0t + (1/2) at2 h = V0t – (1/2) gt2 2h = 2V0t – gt2 gt2 – 2V0t + 2h = 0 t2 – {(2V0) / g}t + (2h / g) = 0 This is quadratic equation in t with roots t1, t2 product of the root = t1t2 = (2h / g)

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