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A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2 find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone ?
from the question, u = 40 m/s g = 10 m/s2 h = ?, v = 0, s = ?
(i) from the formula, v2 = u2 – 2gh
0 = (40)2 – 2 × 10 × h
20h = 1600
h = 1600/20
h = 80m
(ii) Net displacement = 0 (Because the stone returns back to the surface after reaching the maximum height.
(iii) Total distance covered by the stone = 80 m + 80 m
= 160m
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