Question

Velocity at half maximum height for a vertically
projected body is​

Answer 1

Answer:

We know

$h = \frac{ {u}^{2} }{2g}$

So we need v at h/2 .

$2as = {v}^{2} - {u}^{2} \\ \\ -2g( \frac{ {u}^{2} }{4g} ) = {v}^{2} - {u}^{2} \\ \\ {v}^{2} = {u}^{2} - \frac{ {u}^{2} }{2} = \frac{ {u}^{2} }{2} \\ \\ \huge{\orange{v = { \frac{1}{\sqrt 2} }u }}$

Answer 2

As the body is projected vertically the angle made by the body with the horizontal is 90

Hmax = u²/2g

Half of the max height= u²/4g

a=-g

As per the formula

V²=U²+2as

V²=U²-2gu²/4g

V²=2U²-U²/2

V=U/√2

I hope this helps ( ╹▽╹ )