Question

A stone is thrown vertically 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

Heya ❣

u = 40 m/s

As the stone is thrown upward the acceleration due to gravity is to be taken negative.

g = - 10 m/s2

v2 - u2 = 2as

For free fall we can write this equation as,

v2 - u2 =2gh

As the stone reaches the maximum height its final velocity v =0

Thus,

0 - (40)2 = 2× (-10) × h

- 1600 = -20 × h

h = 80 m

So, the maximum height to which the stone can reach is 80 m.

The total distance covered by the stone = 80 + 80 = 160 m

And as th stone comes back to its initial position the displacement of the stone = 0

Answer 2

$\huge\bold\purple{Bonjour!!}$

$\huge\mathfrak{Answer:-}$

⏩ Here, u= 40 m/s, g= -10 m/s²,

At the maximum height, h, final velocity, v= 0

Now, as v² - u² = 2gs,

0² - 40² = 2 × (-10 )h

=> h= 40×40/20 = 80 m. Ans.

Therefore,

→ the total distance covered = h+h = 80 + 80 = 160 m. Ans.

and

→ the net displacement = 80 - 80 = 0. Ans.

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Hope it helps...:-)

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♣ WALKER ♠