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?
To find the height attained by the stone and the time taken to reach that height, we can use the kinematic equations of motion.
The kinematic equation we need to use in this case is:
v^2 = u^2 + 2as
where:
- v is the final velocity of the stone (which will be 0 m/s when it reaches the highest point)
- u is the initial velocity of the stone (which is 5 m/s upwards)
- a is the acceleration of the stone during its motion (which is 10 m/s^2 downwards)
- s is the distance or height attained by the stone (which is what we want to find)
First, let's find the time it takes for the stone to reach its highest point. We know that the final velocity (v) is 0 m/s at the highest point. The initial velocity (u) is 5 m/s upwards, and the acceleration (a) is 10 m/s^2 downwards.
Using the equation v = u + at, we can solve for time (t):
0 = 5 - 10t
Rearranging the equation, we get:
10t = 5
t = 0.5 seconds
So, it will take 0.5 seconds for the stone to reach its highest point.
Now, let's find the height attained by the stone (s) using the same equation:
v^2 = u^2 + 2as
Since the stone reaches its highest point and comes to rest there, the final velocity (v) is 0 m/s. The initial velocity (u) is 5 m/s upwards, and the acceleration (a) is 10 m/s^2 downwards. We need to find the height (s).
0^2 = 5^2 + 2(-10)s
0 = 25 - 20s
20s = 25
s = 1.25 meters
So, the stone will reach a height of 1.25 meters.
V^2 = Vo^2 + 2g*h
h = (V^2-Vo^2)/2g
h = (0-5^2)/-20 = 1.25 m.
V = Vo + g*t
t = (V-Vo)/g = (0-5)/-10 = 0.5 s.