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A stone is thrown vertically upward with an initial velocity of $40\, m/s$. Taking $g = 10 \,m/s^2$, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
$50\,m$ ,$140\, m$ and $5$
$80\,m$ $160\, m$ and $0$
$30\,m$ , $150\, m$ and $3$
$20\,m$ , $170\,m$ and $2$
Solution
According to the equation of motion under gravity $v^2 – u^2 = 2gs$
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, $0^2 40^2 = 2(10)h ⇒ h = 40\times 40 / 20 = 80\,m$
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 $= 80 + (80) = 0$.
