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.
Click to See Solution :
tBC = 2 t
So, for ABC part

tAC = 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
uy = 15
$h = \frac{u_y^2}{2 g} = \frac{225}{20} = \frac{45}{4}$
Maximum height attained $= 20 + \frac{45}{4} = \frac{125}{4} m$