Answer
Please see the work below.
Work Step by Step
(a) We know that
The speed of the first driver:
$v_1^2=v_{\circ}^2-2gh$
We plug in the known values to obtain:
$v_1^2=(1.80)^2-2(9.8)\times (-3.00)$
$v_1^2=62.04$
$v_1=7.87\frac{m}{s}$
Now we can we find the speed of the second driver:
$v_2^2=v_{\circ}^2-2gh$
We plug in the known values to obtain:
$v_2^2=(0.0)^2-2(9.8)(-3.00)=58.8$
$v_2=7.67\frac{m}{s}$
(b) Thus, the first driver hits the water first.
