Question

If a particle is dropped from top of a tower, then ratio of times in falling first three successive equal distances h is

Answer 1

when a particle is dropped from a top of tower , initial velocity of particle , u = 0

time taken to fall h m down , t1

using formula , s = ut + 1/2at²

h = 0 + 1/2 × gt1²

or, t1 = √{2h/g} ......(1)

velocity after time t1

v = u + at

= 0 + g × t1

= g × √{2h/g} = √{2gh}

again Let us consider that time taken to reach h m bellow is t2.

here, initial velocity, u = √{2gh}

using formula, s = ut + 1/2 at²

h = √{2gh}t2 + 1/2gt2²

2h = √{8gh}t2 + gt2²

t2 = {-√(8gh) +√{8gh+8gh}}/2g

t2 = {-√(8gh) + √(16gh)}/2g

= -√{2h/g} + √{4h/g}

from equation (1),

t2 = (-1 + √2)t1 .......(2)

velocity , v = u + at2

v = √(2gh) + g ×(-1 + √2)√{2h/g}

= √{2gh} + (-1 + √2)√{2gh}

= √(4gh)

once again let time taken to reach h m is t3

initial velocity, u = √(4gh)

h = √(4gh)t3 + 1/2 gt3²

2h = √(16gh)t3 + gt3²

t3 = {-√(16gh) + √{16gh+8gh}}/2g

= -√{4h/g} + √{6h/g}

from equation (1),

= (-√2 + √3)t1.......(2)

from equation (1), (2) and (3),

1 : (-1 + √2) : (-√2 + √3)