Question

a stone is thrown vertically upward with an initial velocity of 40m/s taking g=10m/s2find the maximum height reached by the stone.What is the net displacement and total distance covers​

Answer 1

Answer:

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u = 40 m/s

As the stone is thrown upward the acceleration due to gravity is to be taken negative.

g = - 10 m/s2

v2 - u2 = 2as

For free fall we can write this equation as,

v2 - u2 =2gh

As the stone reaches the maximum height its final velocity v =0

Thus,

0 - (40)2 = 2× (-10) × h

- 1600 = -20 × h

h = 80 m

So, the maximum height to which the stone can reach is 80 m.

The total distance covered by the stone = 80 + 80 = 160 m

And as th stone comes back to its initial position the displacement of the stone = 0