If a body is projected vertically up it's velocity decreases to half of its initial velocity at a height 'h' above the ground The maximum height reached by it is

Question

Ans:4h/3

Answer 1

v = Vi - g t
h = Vi t - .5 g t^2

at height h, v = Vi/2
Vi/2 = Vi - g t
g t = Vi/2
t = Vi/(2 g) at height h
h = Vi^2/(2g) - .5g (Vi^2/4g^2)
h = (1/2)Vi^2/g - (1/8)Vi^2/g
h = (3/8) Vi^2/g

at max height H, v = 0
0 = Vi - g T where T is time at max h
T = Vi/g
H = Vi T -.5 g T^2
H = Vi^2/g -.5 g (Vi^2/g^2)

H = .5 [Vi^2/g] = .5 [8h/3) = (8/6)h
= 4h/3