#### Answer

(a) $KE = 6.75\times 10^5~J$
(b) h = 45.9 m
(c) The answer to part (b) does not depend on the mass.

#### Work Step by Step

(a) We can find the kinetic energy of the car;
$KE = \frac{1}{2}mv^2$
$KE = \frac{1}{2}(1500~kg)(30~m/s)^2$
$KE = 6.75\times 10^5~J$
(b) The potential energy at the height $h$ would be equal to the kinetic energy just before impact;
$PE = KE$
$mgh = \frac{1}{2}mv^2$
$h = \frac{v^2}{2g}$
$h = \frac{(30~m/s)^2}{(2)(9.80~m/s^2)}$
$h = 45.9~m$
(c) Since the required height $h$ is equal to $\frac{v^2}{2g}$, the answer to part (b) does not depend on the mass.