## Essential University Physics: Volume 1 (3rd Edition)

(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.