We can use #v=u+at# to calculate speed of the ball after #ts#,

here, #u=4ms^-1,a=g,t=2#

So, #v=4+2*9.8=23.6ms^-1#

And,if its speed is #V# while touching the ground,then we can use, #v^2 =u^2 +2gh#

here, #h=16m#

So, #V =sqrt(16+2*16*9.8)=18.157ms^-1#

Oops! The velocity value while touching the ground is less than the velocity found at #2s#,that means the ball has reached the ground before that.

Let's calculate time required to reach the ground,using #V=u+g t#

So, #t=(18.157-4)/9.8=1.45 s#

So,now if we consider that after the impact the ball went upwards,and also consider that coefficient of restitution was #1#,then that ball will go up with the same velocity.

So,for #(2-1.45)=0.55s# while going up,if it achieves a velocity of #v#,then we cans say,

#v=u-g t#

here, #u=18.157ms^-1#

So, #v=12.767ms^-1# (upwards)(this is the velocity at the end of #2s#)