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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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We know that the force of gravity due to the earth acts on each and every object. When we were holding the stone in our hand, the stone was experiencing this force, but it was balanced by a force that we were applying on it in the opposite direction. As a result, the stone remained at rest. Once we release the stone from our hands, the only force that acts on it is the gravitational force of the earth and the stone falls down under its influence. Whenever an object moves under the influence of the force of gravity alone, it is said to be falling freely. Thus the released stone is in a free fall. In free fall, the initial velocity of the object is zero and goes on increasing due to the acceleration due to the gravity of the earth. During free fall, the frictional force due to air opposes the motion of the object and a buoyant force also acts on the object. Thus, true free fall is possible only in a vacuum.
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The stone held in the hand is stable because on it __.
- two unbalanced forces are exerted.
- only the gravitational force of the earth is exerted.
- gravitational force of the earth is not exerted.
- two balanced forces are exerted.
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