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(B) Using the equation of motion $v^2 - u^2 = 2as$,where $a = -g$ for upward motion.Given:Initial velocity $u = 40 \, m/s$Final velocity at maximum he

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$80 \, m, 160 \, m$ and $0$

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(B) Using the equation of motion $v^2 - u^2 = 2as$,where $a = -g$ for upward motion.Given:Initial velocity $u = 40 \, m/s$Final velocity at maximum height $v = 0 \, m/s$Acceleration $g = 10 \, m/s^2$Using $v^2 = u^2 - 2gh$:$0^2 = (40)^2 - 2(10)h$$20h = 1600$$h = 80 \, m$ (Maximum height).Total distance covered = distance upward + distance downward = $80 \, m + 80 \, m = 160 \, m$.Net displacement = final position - initial position = $0 \, m$ (since the stone returns to the starting point).

$A$ 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?

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