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?
Answers
Answer:
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? Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.
Given:
The initial velocity of the Stone, u = 5 m/s
Final velocity, v = 0 ( Since the stone comes to rest after reaching max. height)
Acceleartion, a = -g = -10m/s² ( - because the stone is thrown upwards against the acceleration)
To find:
The height attained by the stone and the time taken to reach there.
Solution:
Let s be the max. the height attained by the stone in time t.
According to the first equation of motion,
v = u=at
0 = 5 + (-10)t
t = 0.5s
According to the third equation of motion,
v² = u² + 2as
0 = 25 = 2(-10) s
s = 1.25m
Hence, the stone attains a height of 1.25m in 0.5s.