15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = \( 10 m/s^2 \) , find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Given data:

Initial velocity u = 40 m/s

\( g = 10 m/s^2 \)

Max height final velocity = 0

We know that, \( v^2 = u^2 – 2 g s \) [negative as the object goes up]

=> \( 0 = (40)^2 – 2 x 10 x s \) => \( s = \frac{(40 x 40)}{20} \)

Maximum height s = 80 m

Total Distance = s + s = 80 + 80

Total Distance = 160 m

Total displacement = 0 (The first point is the same as the last point)