Question

The average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity. What is the velocity of the body as it reaches the ground?

Answer 1

Howdy!!

your answer is -------

→ A freely falling body has average velocity

V(av.) = (1/2). numerical value of g in m/s

suppose it travelled for t secs

so velocity of striking the ground v = u + g×t = g×t

if it has travelled a distance h in t seconds

h= (1/2) g×t^2

but the same distance it must have travelled with average velocity V(av.)

h = V(av.)× t =(g/2)×t = (1/2) g×t^2

therefore t= 1 s

so, v = g×1 m/s

the velocity with which it strikes the ground = g m/s

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hope it help you

Answer 2

Answer:

The velocity of the body as it reaches the ground is g.

Explanation:

Given that,

The average velocity of a freely falling body is numerically equal to half of the acceleration due to gravity.

$v_{ag} = \dfrac{g}{2}$

Here, v = velocity

g = acceleration due to gravity

The average velocity is

$v_{ag}=\dfrac{u-v}{2}$

Here, u = initial velocity

v = final velocity

Put the value of average velocity

$\dfrac{g}{2}=\dfrac{v}{2}$

Here, initial velocity is zero

$v = g$

Therefore, The velocity is g.

Hence, The velocity of the body as it reaches the ground is g.