Answer:
d) g/2
Explanation:
We need to use one of Newton's equations of motion to find the position of the stone at any time t.
x(t) = x₀(t) + ut - ¹/₂at²
Where
x₀(t) = initial position of the stone.
x(t) - x₀(t) = distance traveled by the stone at any time.
u = initial velocity of the stone
a = acceleration of the stone
t = time taken
On both planets, before the stone was thrown by the astronaut, x = 0 and t = 0.
=> 0 = x₀(t)
=> x₀(t) = 0
On earth, when the stone returns into the hand of the astronaut at time T on earth, x = 0.
=> 0 = 0 + uT - ¹/₂gT² (a = g)
=> uT = ¹/₂gT²
=> g = 2u/T
On planet X, when the stone returns into the hand of the astronaut, time = 2T , x = 0.
=> 0 = 0 + u(2T) - ¹/₂a(2T)²
=> 2uT = 2aT²
=> a = u/T
By comparing we see that a = g/2.
