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A stone is thrown in a vertically upward direction with a velocity of 5 ms~1. If the acceleration of the stone during its motion is 10 ms~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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A stone is thrown in a vertically upward direction with a velocity of ...
Height attained by the stone and time taken to reach there
- Initial velocity and acceleration
The stone is thrown upward with an initial velocity of 5 m/s. The acceleration acting on the stone is 10 m/s^2 in the downward direction.
- Time taken to reach the highest point
To find the time taken to reach the highest point, we can use the equation: v = u + at, where v is the final velocity (0 m/s at the highest point), u is the initial velocity, a is the acceleration, and t is the time taken.
Plugging in the values, we get:
0 = 5 + (-10)t
t = 0.5 seconds
- Height attained by the stone
Using the equation: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time taken.
Plugging in the values, we get:
s = 5(0.5) + (1/2)(-10)(0.5)^2
s = 2.5 - 1.25
s = 1.25 meters
- Conclusion
The stone will reach a height of 1.25 meters and it will take 0.5 seconds to reach that height.
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A stone is thrown in a vertically upward direction with a velocity of 5 ms~1. If the acceleration of the stone during its motion is 10 ms~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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