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 when it falls back to the ground ?
Answers
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 when it falls back to the ground ?
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
ANswer :
- Maximum height attained by the stone = 80m
- Net Displacement = 0 m
- Total Distance covered by stone = 160 m
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).
_________________________
Answer:
80 metres
Explanation:
→ 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