A stone is thrown vertically upwards with a speed of 30 m/s. How high will it go before it begins to fall. Acceleration is 10 m/s² in the downward direction.
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Answered by
4
Given:
- Initial velocity,u = 30 m/s
- Acceleration due to gravity,g = -10 m/s² [ retardation ]
- Final velocity,v = 0 m/s
To be calculated:
How hight will it go before it begins to fall.
Formula used:
v² = u² + 2gh
Solution:
For the freely falling body,
v² = u² + 2gh
★ Putting the values in the above formula,we get
0² = 30² + 2 × -10 × h
⇒ 0 = 900 - 20h
⇒ 20h = 900
⇒ h = 900/20
⇒ h = 45 m
Thus, the stone will go to a maximum height of 45 metres before it begins to fall.
Answered by
2
Answer:
45 metres
Explanation:
Given :
- Initial velocity of the stone = 30 m/s
- Acceleration in upward direction = - 10 m/s² (as +10 m/s² in downward direction)
- Final velocity = 0 m/s (As the velocity at the highest point of the stone will be zero)
To find :
- Maximum height reached by the stone
Using third equation of motion :
V²-u²=2as
0²-30²=2×-10×s
0-900= - 20s
-900= - 20s
S = 900/20
S = 45 metres
The maximum height reached by the stone is equal to 45 metres
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