Physics, asked by uslatha23, 1 year ago

A stone is thrown vertically upward with a speed of 40 m/s.the time interval for which the partical was above 40m from the ground is?(g=10m/s)

Answers

Answered by Anonymous
0

Hey maTe..!! .

Your Answer is..!!

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 = 160m

Net displacement during its upward and downward journey = 80 + (−80) = 0

HOpe iT HelP YOu DEaR

JaI HinD

JaI BhaRat

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