Question

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

Answer 1

Answer:

4h/3

Explanation:

up warard distance = ut - gt²/2

v = u - gt = u/2 , t = u/2g

h = u²/2g - u²/8g = 3/8 × u²/g

max height H = 1/2 u²/g = 4/8 × u²/g = 4h/3

Answer 2

Answer:

4h/3

Explanation:

3rd equation of motion :$v^{2}$ - $u^{2}$ = -2gh

according to the question v = u/2

 [ $\frac{u}{2}$ ]^2 - u^2 = -2gh

$u^{2}$ / 4 - $u^{2}$ = -2gh

-3$u^{2}$ / 4 = -2gh

3$u^{2}$ / 4 = 2gh

$u^{2}$ = $\frac{2gh}{4/3}$ = 8gh / 3

maximum height = $\frac{u^2}{2g}$

                            = (8gh/3)/ 2g

                            = 8h/6

                            =4h/3