An astronaut is on the surface of other planet whose air resistance is negligible. To measure the acceleration due t o gravity `(g)`, he throws a stone upwards. He observer that the stone reaches to a maximum height of `10m` and reaches the surface `4` second after it was thrown. find the accelertion due to gravity `(g)` on the surface of that planet in `m//s^(2)`.

Answer

`5`

Explanation

`(u^(2))/(2g) = 10 m`
`(2u)/(g) = 4sec`
Solving, `g = 5 m//s^(2)`