In the first case, she is initially going upwards at 4m/s, but she will decelarate to 0 and then accelerate to -4m/s as she passes the bar. This is because the ‘g’ is acting over the same distance as she goes up and then as she goes down.

You know that v^2 = u^2 + 2as

‘u’ in the first case is -4, in the second case is -3

‘a’ and ‘s’ are the same in both cases (-g and -h the height of the bar)

v = root( (-3)^2 + 2gh )

w = root( (-4)^2 + 2gh )

w – v = (the difference of the two expressions)

You can see that the final velocity in the first case will have a larger magnitude, but the exact difference depends on the height.