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

According to the equation of motion under gravity v² − u² = 2gs

Where,

u = Initial velocity of the stone = 40 m/s

v = Final velocity of the stone = 0 m/s

s = Height of the stone

g = Acceleration due to gravity = −10 ms⁻²

Let h be the maximum height attained by the stone.

Therefore, 0² − 40² = 2(−10)ℎ ⇒ ℎ = {40 \times 40 \over 20} = 80 𝑚

Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m

Net displacement during its upward and downward journey = 80 + (−80) = 0.