Question

A stone falls freely under gravity. It covers distances 1 h , 2 h and 3 h in the first 3s, the next 3s and the next 3s respectively. The relation between 1 h , 2 h and 3 h is

Answer 1

Explanation:

A stone falls freely under gravity .

so, initial velocity of Stone, u = 0

use formula , s = ut + 1/2at²

then, -h1 = 0 - 1/2g(5)²

h1 = 125m .......(i)

let v is the velocity after 5 second.

then, v² = u² + 2as

v² = 0 + 2(-g)(-125)

v² = 2 × 10 × 125 = 2500

v = -50 m/s {negative sign show body falling downward}

now, stone covered h2 distance in next 5 second .

so, initial velocity of stone in this case ,u' = v = -50m/s

now, use formula, S = ut + 1/2at²

-h2 = -50 × 5 - 1/2 × 10 × 5²

-h2 = 250 + 125 = 375 m ......(ii)

velocity of stone , v' = u' + at

= -50 -10 × 5 = -150 m/s

now, stone covered h3 distance in next 5 second.

so, initial velocity of stone in this case ,u" = v' = -150m/s

use formula , S = ut + 1/2at²

-h3 = -150 × 5 - 1/2 × 10 × (5)²

h3 = 750 + 125 = 875 ......(iii)

from equations (i), (ii) and (iii),

h2 = 3h1 and h3 = 7h1

or, h1 : h2 : h3 = 1 : 3 : 7

Hope it will help you

✌️sai

Answer 2

ut + 1/2at²

then, -h1 = 0 - 1/2g(5)²

h1 = 125m .......(i)

let v is the velocity after 5 second.

then, v² = u² + 2as

v² = 0 + 2(-g)(-125)

v² = 2 × 10 × 125 = 2500

v = -50 m/s {negative sign show body falling downward}

now, stone covered h2 distance in next 5 second .

so, initial velocity of stone in this case ,u' = v = -50m/s

now, use formula, S = ut + 1/2at²

-h2 = -50 × 5 - 1/2 × 10 × 5²

-h2 = 250 + 125 = 375 m ......(ii)

velocity of stone , v' = u' + at

= -50 -10 × 5 = -150 m/s

now, stone covered h3 distance in next 5 second.

so, initial velocity of stone in this case ,u" = v' = -150m/s

use formula , S = ut + 1/2at²

-h3 = -150 × 5 - 1/2 × 10 × (5)²

h3 = 750 + 125 = 875 ......(iii)

from equations (i), (ii) and (iii),

h2 = 3h1 and h3 = 7h1

or, h1 : h2 : h3 = 1 : 3 : 7