## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(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.
(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.