On the earth,when an astronaut throws a 0.250-kg stone vertically upward,it returns to his hand a time T later.On planet X he finds that,under the same circumstances,the stone returns to his hand in 2T.In both cases,he throws the stone with the same initial velocity and it feels negligible air resistance.The acceleration due to gravity on planet X (in terms of g)is
A)g/4.
B)g/2.
C)g/ .
D)g .
E)2g.

Question

On the earth,when an astronaut throws a 0.250-kg stone vertically upward,it returns to his hand a time T later.On planet X he finds that,under the same circumstances,the stone returns to his hand in 2T.In both cases,he throws the stone with the same initial velocity and it feels negligible air resistance.The acceleration due to gravity on planet X (in terms of g)is
A)g/4. 
B)g/2. 
C)g/ . 
D)g . 
E)2g.

Quizplus · Accepted Answer

We can use the equation for the time it takes for an object to reach its maximum height when thrown vertically upward:
$$
T=\frac{2v_0}{g}
$$
where $v_0$ is the initial velocity and $g$ is the acceleration due to gravity.

We know that the initial velocity is the same in both cases, so we can set up a ratio of the times on earth and planet X:
$$
\frac{T_{earth}}{T_{X}}=\frac{2v_0}{g_{earth}}\cdot\frac{g_{X}}{2v_0}=\frac{g_{X}}{g_{earth}}
$$
Solving for $g_X$, we get:
$$
g_X = \frac{T_{X}}{T_{earth}}\cdot g_{earth}
$$
Since the stone takes twice as long to return on planet X, we have:
$$
g_X =\frac{2T_{earth}}{T_{earth}}\cdot g_{earth} = 2g_{earth}
$$
Therefore, the acceleration due to gravity on planet X is twice that of earth, so the answer is (B) $g/2$.

On the Earth,when an Astronaut Throws a 0