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Question
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?
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Solution
Initial velocity (u) = 40m/s
g = −10m/s2
Final velocity at maximum height (v) = 0
Third equation of motion
v2 = u2 + 2gh
0 = 402 + 2 × (−10) × h
1600 = 20h
80m = h
Max Distance = 80m
Total distance travelled by the stone = 80 + 80
= 160m
Because the stone was thrown upwards, but it came back.
Hence, total displacement = 0
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