2 seconds rising and 2 seconds falling
h = (1/2)g t^2 falling
10 = (1/2) g (4) = 2 g
g = 5 m/s^2
An astronaut is on the surface of planet whose air resistance is negligible. To measure the acceleration due to gravity(g) he throws a stone upwards. He observe that the stone reaches the surface to a maximum ht of h =10m(which is negligible is compare radius of planet) and reaches the surface 4 sec after it was thrown. Find the acceleration due to gravity(g) on the surface of that planet
1 answer
1 answer