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?

Given that

Initial velocity u = 40m/s

g = 10 m/s2

Max height final velocity = 0

Find out

The maximum height reached by the stone and the net displacement and the total distance covered by the stone.

Solution

Let us solve using third equation of motion

v2 = u2 – 2gs [negative as the object goes up]

0 = (40)2 – 2 x 10 x s

s = (40 x 40) / 20

Maximum height s = 80m

Total Distance = s + s = 80 + 80

Total Distance = 160m

Total displacement = 0 (The first point is the same as the last point)

Answer

Maximum height s = 80m

Total displacement = 0 

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