#### Answer

(a) The car lands 61 meters from the base of the cliff.
(b) The impact speed is 31 m/s

#### Work Step by Step

(a) We can find the vertical component of velocity when the car lands on the ground. Note that downward is the negative direction:
$v_y^2 = v_{0y}^2+2ay$
$v_y = \sqrt{v_{0y}^2+2ay}$
$v_y = \sqrt{[(20~m/s)~sin(20^{\circ})]^2+(2)(-9.80~m/s^2)(-30~m)}$
$v_y = -25.2~m/s$
We can find the time it takes for the car to land on the ground.
$t = \frac{v_y-v_{0y}}{g}$
$t = \frac{-25.2~m/s-(20~m/s)~sin(20^{\circ})}{-9.80~m/s^2}$
$t = 3.27~s$
We can find the horizontal distance the car travels in this time.
$x = v_x~t$
$x = (20~m/s)~cos(20^{\circ})(3.27~s)$
$x = 61~m$
The car lands 61 meters from the base of the cliff.
(b) We can find the car's impact speed.
$v = \sqrt{(v_x)^2+(v_y)^2}$
$v = \sqrt{[(20~m/s)~cos(20^{\circ})]^2+(-25.2~m/s)^2}$
$v = 31~m/s$
The impact speed is 31 m/s.