Question

$\huge\underline{\bf \orange{Question :}}$

A stone is thrown in a vertically upward direction with a velocity of 5m/s. If the acceleration of the stone during its motion is 10m/s2 in the downward direction , what will be the height attained by the state and how much time will it take to reach there​

Answer 1

Answer:

Explanation:

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 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 x s

- 25/- 20 = s

⇒ s = 1.25

Hence, the height attained by stone is 1.25 seconds.