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?​

Answer 1

Given,

u = 40m/s

g = -10 m/s² ( negative sign shows downward acceleration)

$S_{max}$ = ?

Since we know that at maximum height the stone will have final velocity 0 (because after that it will start descending)

so, v=0

Now, using the equation,

v² = u² + 2as

(0)² = (40)² + 2(-10)s

⇒1600 = 20s

⇒ s = 80m

Now since finally the stone will be on ground i.e. its initial position so its net displacement will be 0

Total distance covered = 2s = 160m

Hope it helps :)

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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