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?

NCERT Class IX
Sciences - Main Course Book

Chapter 10. Gravitation

Answer 1

Maximum height reached by the stone(H) = u²/2g
                                [where u = initial velocity and g = gravitational acceleration.]
                                                         =40²/(2×10).
                                                         = 80 m.
Net displacement = 0 m.[ Initial position coincides with final position.]
Total distance covered = (2×H) = (2×80) = 160 m. (Ans.)

Answer 2

_/\_Hello mate__here is your answer--

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u = 40 m/s

v = 0 m/s

s = Height of the stone

g = −10 ms^−2 ( upward direction)

Let h be the maximum height attained by the stone.

Using equation of motion under gravity

v^2 − u^2 = 2gs

⇒0^2 − 40^2 = 2(−10)ℎ

⇒ ℎ =40×40/ 20 = 80 m

Therefore, total distance covered by the stone during its upward and downward journey

= 80 + 80 = 160 m

Net displacement during its upward and downward journey

= 80 + (−80) = 0

I hope, this will help you.☺

Thank you______❤

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