Question

a body is thrown vertically up with velocity u . it passes three points A B C in its upward journey with velocity u/2 u/3 u/4 respectively . the ratio of separation between point A & B and between B & C .I.e AB/BC is

Answer 1

answer : 20 : 7

A body is thrown vertically upward with speed u.

Let point A is located at s distance from the ground

initial velocity = u

velocity at point A ,v= u/2,

using formula, v² = u² + 2as

or, (u/2)² = u² + 2(-g)s

or, u²/4 - u² = -2gs

or, -3u²/4 = -2gs

or, s = 3u²/8g .....(1)

similarly, velocity at point B, v = u/3

then, (u/3)² = u² + 2(-g)s'

or, -8u²/9 = -2gs'

or , s' = 4u²/9g.....(2)

and velocity at point C, v = u/4

then, (u/4)² = u² + 2(-g)s"

or, -15u²/16 = -2gs"

or, s" = 15u²/32g ....(3)

so, seperation between point A and B , AB = s' - s = (4u²/9g) - (3u²/8g)

= (32 - 27)u²/72g

= 5u²/72g

and seperation between B and C , BC = s" - s'

= (15u²/32g) - (4u²/9g)

= (135 - 128)u²/(288g)

= 7u²/228g

so, AB/BC = (5u²/72g)/(7u²/288g)

= 5 × 288/72 × 7

= 20/7