Question

If the average velocity of a free falling body is numerically equal to half the acceleration due to gravity. Find the velocity of the body as it reaches the ground

SOLUTION REQUIRED


ANSWER IS
$g \times \sqrt{2}$

Answer 1

g×√2= the gravitational force × the root of to balance the eq to make a perfect velocity

Answer 2

Answer: The velocity of the body as it reaches the ground is g m/s.

Explanation:

Given that,

Acceleration due to gravity =$\dfrac{g}{2}$

We know that,

The average velocity is the displacement upon total time.

$v = \dfrac{D}{t}$

Now, using second equation of motion

$v = \dfrac{ut-\dfrac{1}{2}gt^2}{t}$

Initial velocity u = 0

$v = \dfrac{-\dfrac{1}{2}gt^2}{t}$

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

Therefore, the time will be

$\dfrac{-\dfrac{1}{2}gt^2}{t}=\dfrac{g}{2}$

$t = -1 s$

Again using equation of motion

$v = u-gt$

$v = 0-g\times(-1)$

$v = g\ m/s$

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