A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g=10m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Given : Initial velocity of stone u = 40m/s
Let the maximum height reached and time taken to reach that height be H and t respectively.
Velocity of the stone at maximum height is zero i.e v = 0
Using, v² - u² = 2aH
where a = -g = -10m/s²
0 - (40)² = 2(-10)H
=> H = 80m
The stone, after reaching the maximum height, starts to fall down and reach the surface again.
∴ Net displacement of the stone is Zero
Total distance covered by the stone S = 2H = 2 × 80 = 160m