→ 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).
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Initial velocity, u=40m/s
Acceleration due to gravity, g=10m/s^2
But because the stone is originally thrown upward the acceleration due to gravity would act in the direction opposite to its motion. So we will use negative signs for g.
i.e., g=−10m/s^2
From the third equation of motion, we have:
v²−u²=2as
for our fr.ee falling stone we can write it as,
v²−u²=2gh
where g is acceleration due to gravity,h is the height of the stone.
Refer to the attachment after this.
