Answer:
Explanation:
Let the ball is thrown upwards with speed u and reaches upto height h.
Use III equation of motion
[tex]v^{2}=u^{2}-2gh[/tex]
here, final velocity v = 0
So, [tex]u = \sqrt{2gh}[/tex]
Now as the ball reaches the maximum height, it starts falling freely. Let it strkes with the ground with velocity v'.
[tex]v'^{2}=u^{2}-2gh[/tex]
here, u = 0
So, [tex]v' = \sqrt{2gh}[/tex]
It means the velocity is same as the velocity of projection.