Question

A particle is released from a tower of height 3h.The ratio of times to fall equal heights h,i.e,t1:t2:t3 is ?

Answer 1

A particle is released from a tower of height 3h. initial velocity of particle , u = 0

using formula , s = ut + 1/2 at²

given time taken to fall first h distance is t1.

so, h = 0 + 1/2 (g)t1²

or, t1 = √{2h/g} .....(1)

velocity of particle after time t1, v = u + at

or, v = 0 + gt1 = g√{2h/g} = √(2gh)

now, again particle is falling downward and time taken to fall next h distance is t2.

so, h = √(2gh)t2 + 1/2 gt2²

or, gt2² + 2√(2gh)t2 - 2h = 0

or, t2 = {-2√(2gh) ± √{8gh + 8gh}}/2g

= {-2√(2gh) + (2√2)√(2gh)}/2g

= √(2h/g) [-1 + √2]

velocity after time t2 , v' = v + gt2

or, v' = √{2gh}+g × √(2h/g) [-1 + √2]

= √(4gh)

similarly, h = √(4gh) t3 + 1/2gt3²

t3 = {-2√(4gh) + √(16gh+ 8gh)}/2g

= √{2h/g} [ -√2 + √3 ]

so, t1 : t2 : t3 = 1 : (-1 + √2) : (-√2 + √3)