#### Answer

$ 56.7m$

#### Work Step by Step

Total energy of the car (kinetic + potential) is: $E_{tot}=E_{kin}+E_{pot}=mgh+\dfrac {mv^{2}_{0}}{2}(1)$
We know that the total energy of the car is conserved. Therefore, at the last hill, the car will climb until its speed become zero. At this time, the kinetic energy of the car will be zero (at highest point) too so at that point, the car will only have potential energy. Now, lets write the total energy at that time:
$E_{tot}=E_{kin}+E_{pot}=mgD+\dfrac {mv^{2}}{2}$
Since $v=0$,
$E_{tot}=mgD\left( 2\right) $
So equating (1) and (2), we get $D=h+\dfrac {v^{2}_{0}}{2g}$
We then substitute the values into this formula to find $D$: $D=h+\dfrac {v^{2}_{0}}{2g}\approx 56.7m$