The velocity of the ball when it strikes the ground, given the data is 21.56 m/s
<h3>Data obtained from the question</h3>
From the question given above, the following data were obtained:
- Time to reach ground from maximum height (t) = 2.2 s
- Initial velocity (u) = 0 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Final velocity (v) =?
<h3>How to determine the velocity when the ball strikes the ground</h3>
The velocity of the ball when it strikes the ground can be obtained as illustrated below:
v = u + gt
v = 0 + (9.8 × 2.2)
v = 0 + 21.56
v = 21.56 m/s
Thus, the velocity of the ball when it strikes the ground is 21.56 m/s
Learn more about motion under gravity:
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Answer:
Explanation:
Let h be the height .
initial velocity in first case u = 0
final velocity v = 6 m /s
acceleration due to gravity g = 9.8 m /s²
v² = u² + 2 g h
6² = 0 + 2 x 9.8 x h
h = 1.837 m .
For second case u = 3 m /s
v² = u² + 2 gh
= 3² + 2 x 1.837 x 9.8
= 9 + 36
= 45 m
v = 6.7 m /s
1 kg of water = 1 L = 1 dm³
762 g of water = 762 cm³
Answer:
389.78681 K
Explanation:
= Initial pressure = 55.1 mmHg
= Final pressure = 1 atm = 760 mmHg
= Boiling point
= Initial temperature = 35°C
= Heat of vaporization = 32.1 kJ/mol
From the Clausius-Claperyon equation
The normal boiling point of the substance is 389.78681 K
