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Question

Question 15
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

According to the third equation of motion under gravity:
v2u2=2 gh
Where,
u = Initial velocity of the stone
v = Final velocity of the stone
h = Displacement of the stone
g = Acceleration due to gravity

We know that velocity at the highest point is equal to zero.
Here, u = 40 m/s, v = 0, g = =10 ms2

Therefore,
0(40)2=2×(10)×h
h=40×4020=80 m
Maximum height reached = 80 m

Therefore, total distance covered by the stone during its upward and downward journey = 80 m + 80 m = 160 m

Net displacement of the stone during its upward and downward journey
= 80 m + (- 80 m) = 0

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