A stone is thrown vertically upward with an initial velocity of 40ms-1. Thanking g=10 ms-2,the maximum displacement and distance covered by the stone on reaching the ground is
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Given -
Stone's Velocity - 40m/s
Stone's acceleration - -10m/s^2(by taking downward direction as negative)
Final Velocity - 0m/s
Answer -
LET THE Maximum Displacement COVERED BE X
2as= v^2 - u^2
2 × -10× X = (0 ^2) - (40^2)
2 × -10 × X = -1600
-20X = -1600
X = -1600/-20
X = 80
Maximum Displacement is 80m
Total Distance covered is -
80m upwards + 80m Downwards
= 160m
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Stone's Velocity - 40m/s
Stone's acceleration - -10m/s^2(by taking downward direction as negative)
Final Velocity - 0m/s
Answer -
LET THE Maximum Displacement COVERED BE X
2as= v^2 - u^2
2 × -10× X = (0 ^2) - (40^2)
2 × -10 × X = -1600
-20X = -1600
X = -1600/-20
X = 80
Maximum Displacement is 80m
Total Distance covered is -
80m upwards + 80m Downwards
= 160m
Please mark as Brainliest
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