Verified answer

Using law, v^(2) - u^(2) = 2as, we can easily find out the answer.

v=?

u=7.3m/s

a=-9.81m/s2

s=0 (displacement is 0)

We find that the initial velocity, and the final velocity is equal, no doubt.

Let intial vel. = u(7.3)

now find the maxm. Height reached by ball

by formula

v^2 = u^2 + 2(-g)h

Where v(final vel.) is zero bcoz at maxm. Height ball will stop. And i used "-g" bcoz gravity is acting downward.

After solving we get h= u^2/2g

now consider the motion of ball going downward from height "h". So now intial velocity is zero bcoz ball starts from rest and we have to find final velocity. Now by applying same formula

(final vel.)^2 = (intial vel.)^2+2(-g)(-h).

I use "-h" bcoz ball is going downward.

Now put the value of "h" in last eqn. From there we get

final velocity = u =7.3 m/s.

So ball will hit ur hand with same speed as u have thrown it up if there is no air resistance.

If you neglect the air friction, then the mechanical energy is conserved, so the speed of the ball as it reaches the earth will be the same, 6 m/s (well, of course, it's directed downwards!).