$A$ stone is thrown vertically upwards with a velocity of $40 \, m s^{-1}$ and is caught back. (Taking $g = 10 \, m s^{-2}$) Calculate the maximum height reached by the stone. What is the net displacement and total distance covered by the stone?
(N/A) Given initial velocity $u = 40 \, m s^{-1}$,acceleration due to gravity $g = -10 \, m s^{-2}$ (acting downwards),and final velocity $v = 0 \, m s^{-1}$ at the maximum height.
Using the equation of motion $v^2 - u^2 = 2as$:
$0^2 - (40)^2 = 2 \times (-10) \times h$
$-1600 = -20 \times h$
$h = 80 \, m$.
Therefore,the maximum height reached is $80 \, m$.
The total distance covered is the sum of the upward and downward journey: $80 \, m + 80 \, m = 160 \, m$.
The net displacement is the difference between the final and initial position. Since the stone returns to the starting point,the net displacement is $0 \, m$.
Using the equation of motion $v^2 - u^2 = 2as$:
$0^2 - (40)^2 = 2 \times (-10) \times h$
$-1600 = -20 \times h$
$h = 80 \, m$.
Therefore,the maximum height reached is $80 \, m$.
The total distance covered is the sum of the upward and downward journey: $80 \, m + 80 \, m = 160 \, m$.
The net displacement is the difference between the final and initial position. Since the stone returns to the starting point,the net displacement is $0 \, m$.
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