Physics, asked by lellasujatha1024, 8 months ago

9. A stone is thrown in a vertically upward direction with a velocity of 5 m s-1. If

the acceleration of the stone during its motion is 10 m s–2 in the downward

direction, what will be the height attained by the stone and how much time

will it take to reach there?​

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Answered by HarshChaudhary0706
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Answer:

Explanation:

Given :-

Initial velocity, u = 5 m/s

Acceleration, a = - 10 m/s²  (As stone is coming downward)

Final velocity, v = 0 (As stone thrown upwards)

To Find :-

Time taken, t = ??

Distance covered, s = ?

Formula to be used :-

1st equation of motion, v = u + at

3rd equation of motion, v² - u² = 2as

Solution :-

Putting all the values, we get

v = u + at

⇒ 0 = - 5 + (- 10) × t

⇒ - 5 = - 10 t

⇒ 5/10 = t

⇒ t = 1/2

⇒ t = 0.5 seconds

Hence, the time taken by stone to reach there is 0.5 seconds.

Now, using v² - u² = 2as

(0)² - (5)² = 2 × (- 10) × s

⇒ - 25 = - 20 × s

⇒ - 25/- 20 = s

⇒ s = 1.25

Hence, the height attained by stone is 1.25 seconds.

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