Question

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? 

Answer 1

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, 0 ² − 40² = 2(−10)ℎ ⇒ ℎ = (40×40) / 20 = 80  

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.