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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Answer 1

Answer:

max height reached is 80m

net displacement is zero

total distance covered is 160m

Answer 2

It is given that,

The initial velocity of the stone, u = 40 m/s

We need to find the maximum height reached by the stone, the net displacement and the total distance covered by the stone.

At maximum height, final velocity, v = 0

Using the equation of kinematics as :

d is the maximum height reached

a = -g

d = U²/2g

d = (40)²/2×10

d = 80 m

As the stone goes up and come down. It means that the distance covered by the stone is 2d i.e. 160 m. For this, the displacement of the stone is 0 as the initial position is equal to the final position.

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