Answer:

• Maximum height attained by the stone = 80m
• NET DISPLACEMENT = 0 m
• Total Distance covered by stone = 160 m

Explanation:

<U>Given:-

• Initial velocity ,u = 40m/s
• Final velocity ,v = 0m/s (as the stone reached its maximum height)
• Acceleration due to gravity ,g = 10m/s²

To FIND:-

• Maximum height ,h
• Net Displacement ,S
• Total distance covered by ball , d

Solution:-

Firstly we calculate the maximum height attained by the stone when it is thrown . Using 3rd equation of motion

• v² = u² + 2gh

where,

• v denote final velocity
• u denote initial velocity
• g denote acceleration due to gravity
• h denote maximum height attained by stone

Substitute the value we get

→ 0² = 40² + 2 (-10) × h

→ 0 = 1600 -20h

→ -1600 = -20h

→ 20h = 1600

→ h = 1600/20

→ h = 80 m

• Hence, the maximum height attained by the stone is 80 metres

The total distance covered by the stone when it falls back to the ground is 80 + 80 = 160 metres

And the net displacement of the stone is 0 metres (as the stone reached its initial position).