Here we have,
Initial velocity (u) = 5 m/s
Final velocity (v) = 0 m/s
Acceleration (a) = – 10 m/s2
Height, i.e. Distance, s =?
Time (t) taken to reach the height =?
We know that, v2 = u2 + 2as
⇒ 0 = (5)2 + 2 × −10 × s
⇒ 0 = 25 − 20s
⇒ s = 25/20 m
⇒ s = 1.25 m
Now, we know that, v = u + at
⇒ 0 = 5 + (–10) × t
⇒ 0 = 5 − 10t
⇒ t = 5/10 s
⇒ t = 0.5 s
Thus, stone will attain a height of 1.25 m and time taken to attain the height is 0.5 s.
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car? 
(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?