Question

A particle is thrown in upward direction with initial velocity V₀. It crosses point P at height h at time t₁ and t₂ so t₁t₂ = –––––– 

(A) (2h / g) 

(B) [(V₀²) / 2g] 

(C) [(2V₀²) / g] 

(D) (h / 2g)

Answer 1

Correct option: (A) (2h / g)

Explanation:

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)