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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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.