Question

a stone is thrown vertically upward with initial velocity 20m/s.taking 'g' as -10m/s2, find maximum height reached by the stone what is the net displacement and the total distance covered by the stone

Answer 1

initial velocity = 20 m/s

g = 10 m/s^2 ( downward)

$maximum\ height = h \\ \\ h = \frac{ {u}^{2} }{2g} = \frac{ {20}^{2} }{2 \times 10} = \frac{20 \times 20}{20} = 20 \: m$
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if we consider only the upward motion,
net displacement = 20m
distance travelled = 20 m
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The ball will fall down to ground. If we consider the downward motion also, the ball will reach it's starting point.

net displacement = 0

total distance covered = 2×h = 20×2 = 40m

Answer 2

_/\_Hello mate__here is your answer--

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u = 20 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 − 20^2 = 2(−10)ℎ

⇒ ℎ =20×20/ 20 = 20 m

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

= 20 + 20 = 40 m

Net displacement during its upward and downward journey

= 20 + (−20) = 0

I hope, this will help you.☺

Thank you______❤

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