Answer
The sandbag's speed when it hits the ground is 30.0 m/s
Work Step by Step
When the sandbag is released, it will have an initial velocity of $10.0~m/s$ upward. We can find the velocity when the sandbag hits the ground:
$v_f^2 = v_0^2+2a\Delta y$
$v_f = \sqrt{v_0^2+2a\Delta y}$
$v_f = \sqrt{(10.0~m/s)^2+(2)(-9.80~m/s^2)(-40.8~m)}$
$v_f = -30.0~m/s$
The sandbag's speed when it hits the ground is 30.0 m/s
