A ball is thrown vertically upwards with a velocity of 6 ms-2. Calculate the maximum height to which it rises. (take acceleration due to gravity = 10 ms-2)
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Answers
Given that, a ball is thrown vertically upward with a velocity of 6 m/s. And. acceleration due to gravity is 10 m/s².
We have to find the maximum height to which the ball rises means 's'.
From the above data, we have initial velocity of the ball i.e. u is 6 m/s, final velocity of the ball i.e. v is 0 m/s and acceleration due to gravity i.e. g is 10 m/s².
As ball is moving against the gravity. So, the value of g will be negative i.e. -10 m/s².
Using the Third Equation Of Motion,
v² - u² = 2gs
Substitute the known values
→ (0)² - (6)² = 2 × (-10) × s
→ 0 - 36 = -20s
→ -36 = -20s
→ 36 = 20s
Divide with 20 on both sides,
→ 36/20 = 20s/20
→ 1.8 = s
Therefore, the maximum height to which the ball rises is 1.8 m.
Given
- Final velocity v = 0 m/s
- Initial velocity u = 6 m/s
- Acceleration due to gravity g = -10 m/s
☯ g = -10 m/s as the body is thrown vertically upwards
To find
- Maximum height h
Solution
- Using third equation of motion :
- v² - u² = 2gs
- 0² - 6² = 2 × (-10)s
- -36 = -20s
- 36 = 20s
☯ Dividing 20 from both sides :
- 36/20 = 20s/20
- 1.8 = s
☯ ∴ Maximum height to which the ball rises is 1.8 m