Question

a stone is thrown vertically in upward direction qith innitial velocity 40m/s taking g=10m/ s find the maximum height and total distance covered by the stone when it falls back to the ground

Answer 1

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 = 160 m Net displacement during its upward and downward journey = 80 + (−80) = 0.Read more on Sarthaks.com - https://www.sarthaks.com/9590/a-stone-is-thrown-vertically-upward-with-an-initial-velocity-of-40-m-s

Answer 2

max hight = 40×4 - 5×16= 80m

total distance =2 × maxm hight

=2×80m=160