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Grade: 11

                        

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

Answers : (1)

Vikas TU
12276 Points
							
let a be the acceleration then,
maximum height = 10 = u^2/2a..............(1)
From newtns laws of eqn.
T = 2u/a = 4
or
u = 2a...............(2)
substituting u in eqn. (1)
2a = 10
a = 5 m/s^2.
 
4 years ago
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