Question

from an elevated point 'A', a stone is projected vertically upwards when stone reaches a distance 'h' below A, its velocity is double of what it was at height 'h' above A. show that greatest height attained by stone is5/3 h

Answer 1

Given :
Distance=h
Initial velocity of stone =u

Velocity at point B , "h" above A is :
Vb²=u²-2gh
Vb=√(u2-2gh)

velocity at point C below "h"
Vc²=u2-2g(-h)

vc²=u2+2ghvc
=√(u2+2gh)

But according to given :

Vc²=4Vb²

(√(u1+2gh)²=4√(u2-2gh)²

u²+2gh=4u²-8gh

u²=10gh/3

Maximum height attained by stone:

Hmax=u²/2g

=10gh /3x2g=5h/3

∴Maximum height attained by stone is 5h/3

Answer 2

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