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?

Open in App

Solution

According to the equation of motion under gravity:

v² − u² = 2 gs

Where,

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

v = Final velocity of the stone = 0

s = Height of the stone

g = Acceleration due to gravity = −10 m s⁻²

Let h be the maximum height attained by the stone.

Therefore,

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

Net displacement of the stone during its upward and downward journey

= 80 + (−80) = 0

Suggest Corrections

Similar questions

g = 10 m / s^{2}

40 m s^{- 1}

g = 10 m s^{- 2}

40 m s^{- 1}

10 m s^{- 2}

Join BYJU'S Learning Program