A stone is thrown vertically upward with an initial velocity of 40 m/s
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?
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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.
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