Question

a body is thrown vertically up with certain velocity if h is the maximum height reached by its position when its velocity reduces to 1 by 3 of its velocity of projection is at​

Answer 1

answer : 8h/9 from ground , h/9 from maximum height.

explanation : Let initial velocity of body is u

given, maximum height reached by body is h.

as we know, at maximum height , velocity of body becomes zero.so, v = 0

now applying formula, v² = u² + 2as

or, 0 = u² + 2(-g)(h)

or, u² = 2gh

or, u = √(2gh).......(1)

let velocity of body reduces to 1/3rd of its velocity of projection at height , y.

so, velocity of body at y , v = u/3 = √(2gh)/3

applying formula, v² = u² + 2as

here, v = √(2gh)/3, u = √(2gh) , a = -g and s = y

so, {√(2gh)/3}² = (√2gh)² + 2(-g)y

or, 2gh/9 - 2gh = -2gy

or, 2g[h/9 - h] = -2gy

or, 8h/9 = y

hence, height of body from the ground is 8h/9 and (h - 8h/9) = h/9 from the maximum height.

Answer 2

Answer:

ans is ...8h/9..........