Question

a stone is thrown vertically upward with an initial velocity of 40 metre per second taking G is equal to 10 M per second find the maximum height reached by the stone what is the net displacement and total distance covered by the stone.please give me full answer.

Answer 1

Hii dear here is your answer......

This answer is verified ✅✅

u = 40 m/s
v = 0 m/s
g = -10 m/s^2
h = ?

${v}^{2} = {u}^{2} + 2gh \\ {0}^{2} = {40}^{2} + 2 \times (- 10) \times h \\ 0 = 1600 - 10h \\ 10h = 1600 \\ h = \frac{1600}{10} \\ h = 160m$
v = u + gt
0 = 40 + (-10)× t
0 = 40 - 10t
10t = 40
t = 40/10
t = 4 s

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

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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