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 10 m/s in the downward direction. What will be the height attained by the stone and how much time will it take to reach there?
Answers
Answer:
v^2 - u^2 = 2gh
0^2 - 5^2m/s = 2*-10m/s^2 *h
-25m/s = -20h
h = 1.25m
t = v - u / g
= 0 - 5m/s / -10m/s^2
= 0.5s
hope it helps
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 0.5 seconds.