A stone is thrown vertically upward with an initial velocity of 40m/s.taking g=10m/s2,find the maximum height reached by the stone.what is the net displacement and the total distance covered by the stone?
Answers
Answer:
Maximum height reached= S
By kinematics equation,
v^2-u^2=2aS
At the highest point, v=0
0-40^2= 2×-10×S
-1600/-20=S
S=80m
Displacement will be zero as the body comes back to initial position downwards after reaching the highest point...(0m)
Distance will be twice of maximum height.. as the object moves upwards then comes down.
2S= Distance
= 2×80= 160m...
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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