#### Answer

The speed at the bottom of the hill is 70.9 km/h. Since this speed is over the speed limit of 70 km/h, it is possible that the police officer will give us a speeding ticket.

#### Work Step by Step

We can convert the initial speed to units of m/s
$v = (35~km/h)(\frac{1000~m}{1~km})(\frac{1~hr}{3600~s})$
$v = 9.7~m/s$
We can use conservation of energy to find the speed at the bottom of the hill.
$KE_f+PE_f = KE_0+PE_0$
$\frac{1}{2}mv_f^2+0=\frac{1}{2}mv_0^2+mgh$
$v_f^2=v_0^2+2gh$
$v_f= \sqrt{v_0^2+2gh}$
$v_f= \sqrt{(9.7~m/s)^2+(2)(9.80~m/s^2)(15~m)}$
$v_f = 19.7~m/s$
We can convert the final speed to units of km/h
$v = (19.7~m/s)(\frac{1~km}{1000~m})(\frac{3600~s}{1~hr})$
$v = 70.9~km/h$
The speed at the bottom of the hill is 70.9 km/h. Since this speed is over the speed limit of 70 km/h, it is possible that the police officer will give us a speeding ticket.