Question

A ball comes down with an increasing speed from the top of a building . Just before it hits the ground , its speed is given by:
${v}^{2} = 4gh \\ {v}^{2} = 2gh \\ v = 2gh \\ {v}^{2} = gh$

Answer 1

v^2 = 2gh is right answer

It comes from the equation, v^2 = u^2 + 2as

where, final velocity becomes zero i.e u=0

    acceleration because of gravitation and distance in height

Thus, a=g and s= h

Answer 2

Answer:

The speed of the ball, just before it hits the ground is given by

$v^2=2gh$.

Explanation:

Assuming, the height of the building is $h$.

The ball is coming down with the increasing speed so if the ball starts coming down from rest then its initial speed is 0.

$u = 0\ m/s$.

The acceleration that acts on the ball when it is falling is the acceleration due to gravity, therefore,

$a = g$.

Now, using the equation,

$v^2-u^2 = 2as\\\text{where,}\ s =h \ \text{and}\ v \ \text{is the final velocity of the ball, just before it hits the ground.}\\\\v^2-0^2=2gh\\v^2=2gh.$