Astone 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 netdisplacement and the total distance covered by the stone?

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Solution

Accordingto the equation of motion under gravity:
v²âˆ’ 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 = âˆ’10 ms^âˆ’2
Leth be themaximum height attained by the stone.
Therefore,
Therefore,total distance covered by the stone during its upward and downwardjourney = 80 + 80 = 160 m
Netdisplacement of the stone during its upward and downward journey
= 80 +(âˆ’80) = 0

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Astone 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 netdisplacement and the total distance covered by the stone?

Astone 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 netdisplacement and the total distance covered by the stone?