Question

A stone is thrown vertically upward direction with the velocity of 5m/s. If the acceleration of the stone during it's motion is 10m/s2 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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Answer 1

Answer:

Given, initial velocity, u = 5ms^−1

Final velocity, v = 0

Since, u is upward & a is downward, it is a retarded motion.

∴a = −10ms^−2

Height attained by stone, s=?

Time take to attain height, t=?

(i) Using the relation, v^2 − u^2 = 2as, we have

s = v^2 − u^2 / 2a

= (0)^2 − (5)^2 / 2 × (−10) = 1.25m

(ii) Using the relation, v = u + at

0 = 5 + (−10)t or

t = 5 / 10 = 0.5s

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Answer 2

Answer:

s= 1.25 m

t= 0.5 s

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 = ?

We know:

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 m.