Given that
Initial velocity u = 40m/s
g = 10 m/s2
Max height final velocity = 0
Find out
The maximum height reached by the stone and the net displacement and the total distance covered by the stone.
Solution
Let us solve using third equation of motion
v2 = u2 – 2gs [negative as the object goes up]
0 = (40)2 – 2 x 10 x s
s = (40 x 40) / 20
Maximum height s = 80m
Total Distance = s + s = 80 + 80
Total Distance = 160m
Total displacement = 0 (The first point is the same as the last point)
Answer
Maximum height s = 80m
Total displacement = 0