Question

a stone is thrown vertically upwards with an initial velocity of 40m/s.Find the
maximum height reached by it.What is net displacement and distance travelled?

Answer 1

Final Answer : h(max) = 80m
Net displacement , s = 0
Distance travelled = 2h(max)= 160m


Solution,
At Max height,
,Final velocity v = 0
Initial velocity u = 40m/s
acceleration, a = -10m/s^2
Now,
using Newtons law of equation,
${v}^{2} - {u}^{2} = 2as$
=>
${0}^{2} - {40}^{2} = 2 \times - 10 \times h(max) \\ = > h(max) = 80m$


Since, Stone is thrown vertically upwards,
and it reaches the same point after thrown.
So, Net Displacement = 0.

Distance travelled = h(max) (for up) + h(max) (for down) = 80m +80m =160m

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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