Question

A stone is thrown vercally upwards with a velocity of 40 m/s and is caught back. Taking g = 10m/s~ calculater the maximum height reached by the stone. What is the net displacement and total distance covered by the stone?

Answer 1

h max-80m
since hmax is equal to square of velocity divided by 2g

Answer 2

_/\_Hello mate__here is your answer--

____________________

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______❤

_______________________❤