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A stone is thrown vertically upward with an initial velocity o... - Science
A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 \text{ m/s}^2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Given:
Initial velocity (
u) =40 \text{ m/s}Final velocity (
v) =0 \text{ m/s}(at maximum height)Acceleration due to gravity (
g) =-10 \text{ m/s}^2(upward motion)
(i) Maximum Height (h):
Using the equation: v^2 - u^2 = 2gh0^2 - (40)^2 = 2 \times (-10) \times h-1600 = -20hh = \frac{1600}{20}h = 80 \text{ m}
(ii) Net Displacement:
Since the stone returns to the starting point, the net displacement is Zero (0).
(iii) Total Distance Covered:
Total distance = Height of ascent + Height of descent
Total distance =80 + 80
Total distance = 160 \text{ m}