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)
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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
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