Answer
37.6 m/s.
Work Step by Step
The rock has a constant downward acceleration. We know the displacement of zero and the time of 6.00 s. Use the kinematics formulas for constant acceleration to find the initial velocity.
Use $y-y_o=v_{oy}t+0.5at^2$
$0= v_{oy}(6.00s)+0.5(-9.80m/s^2)(6.00s)^2$
$v_{oy}=29.4m/s$
Now that we have the initial velocity, find the final speed.
$v_f^2=v_o^2+2a(\Delta x)$
$v_f^2=(29.4m/s)^2+2(-9.8m/s^2)(-28.0m)$
$v_f= 37.6 m/s$
