(2) s = ut + 1/2*at^2
(3) v^2 + u^2 = 2as
In this problem, you just have to identify when to use which equation.
(a) final speed = v = 0 [because at the highest point of motion, the stone will momentarily come to rest.]
(b) initial speed = u = 20
final speed = v = 0
acceleration = g = -9.8
Explanation for -ive sign: velocity and acceleration are vector quantities, so they must show us magnitude as well as direction. I'm assuming the upward direction to +ive and downward to be negative. Then, as g acts in downward direction, so we put the minus sign.
Now, use equation (1) to find the value of t. Do calculations yourself.
(c) After getting 't' , put it in equation (2) and you'll get the required height, 's'.
PART (d)
Now, forget everything done above and solve this part independently.
For convenience, we assume the downward direction to be positive.Then,
initial velocity = u = 0 [as it is at momentarily at rest at the highest point]
Final velocity = v = 50
acceleration= g = 9.8
put them in equation (3 ) to get the overall height of the cliff.