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An astronaut is on the surface of the planet whose air resistance is negligible to measure acceleration due to gravity he throws a stone vertically upward. he found that the stone reaches to a maximum height of 10 m and reaches the surface 4 sec after it was thrown. find the acceleration due to gravity on the surfaces of that planet in m/s^2
An astronaut is on the surface of the planet whose air resistance is negligible to measure acceleration due to gravity he throws a stone vertically upward. he found that the stone reaches to a maximum height of 10 m and reaches the surface 4 sec after it was thrown. find the acceleration due to gravity on the surfaces of that planet in m/s^2

```
4 years ago

Vikas TU
14149 Points
```							let a be the acceleration then,maximum height = 10 = u^2/2a..............(1)From newtns laws of eqn.T = 2u/a = 4oru = 2a...............(2)substituting u in eqn. (1)2a = 10a = 5 m/s^2.
```
4 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions