HI
I am having problems understanding this question, if someone could help me as soon as possible, it will be much appreciated.

A stunt man drives a car at a speed on 20 m/s off a 30 m high cliff. The road
Leading to the cliff is inclined up ward at an angle of 20°.
(I) How far from the base of the cliff does the car land?
(ii) What is the car’s impact speed?

Sorry... hopefully, that's not too late..

First, have a sketch; considering the vertical component, your initial speed will be 20 sin(20). Use

s=ut+\frac{1}{2}at^2 to find the time taken for the car to reach the bottom of the cliff, displacement s being 30m, initial velocity u being 20 sin(20) m/s and acceleration a being 9.81m/s^2.

Then, to find the car final speed, you have to add the vertical component of the speed and its horizontal component. Your horizontal being constant at 20 cos(20) m/s, you have to find the final vertical speed. You can use

v^2 = u^2 + 2as (u=20sin(20) m/s, a=9.81m/s^2, s=30m)

or use the time you got before and using the formula

v=u+at

however, if you made a mistake above, then using v = u+at will be wrong. Anyway, if you id it right, you should have the same answer with both formulae.

Now, using Pythagoras' Theorem, do the square root of the sum of the square of the vertical and horizontal components,

\sqrt{(20cos(20))^2+(v^2)}

There! :) Hope it helped!

I think I don't know

Sigh, then what the use of posting, arjun?