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
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
To find the acceleration due to gravity (g) on the surface of the planet, we can use the equation of motion for vertical motion:
h = (1/2)gt^2 + v₀t + h₀
Where:
h is the maximum height reached by the stone (10 m)
g is the acceleration due to gravity
t is the time taken for the stone to reach the surface (4 s)
v₀ is the initial velocity (which is 0 in this case)
h₀ is the initial height (also 0 in this case, because we are measuring from the surface of the planet)
Plugging in the known values, the equation becomes:
10 = (1/2)g(4)^2 + 0 + 0
Simplifying the equation:
10 = 8g
To solve for g, divide both sides of the equation by 8:
g = 10/8
g ≈ 1.25 m/s²
Therefore, the acceleration due to gravity (g) on the surface of the planet is approximately 1.25 m/s².