Question

a stone is thrown vertically upwards with an intial velocity of 40m/s .tosing g 10m/s find its maximum height reached by the stone what is the net displacement and the total distance covered by the stone​

Answer 1

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.

Answer 2

v2 − u2 = 2 gs

Whe

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 = −10 m s−2

Let h be the maximum height attained by the stone.

Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m

Net displacement of the stone during its upward and downward journey

= 80 + (−80) = 0

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