Q: A particle is projected from the ground level. It just passes through upper ends of vertical poles A, B, C of height 20 m, 30 m and 20 m respectively. The time taken by the particle to travel from B to C is double of the time taken from A to B. Find the maximum height attained by the particle from the ground level.

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_{AB}= t

t

_{BC}= 2 t

So, for ABC part

t_{AC} = 3 t $= \frac{2 u_y}{g}$

$u_y = \frac{3}{2}g t $

Also ; $10 = u_y – \frac{1}{2}g t^2 = g t^2 $

t = 1 s

u_{y} = 15

$h = \frac{u_y^2}{2 g} = \frac{225}{20} = \frac{45}{4}$

Maximum height attained $= 20 + \frac{45}{4} = \frac{125}{4} m$