Question

Suppose you are an astronaut and you have been stationed on a distant planet. you would like to find the acceleration due to the gravitational force on this planet so you devise an experiment. you throw a rock up in the air with an initial velocity of 11 m/s and use a stopwatch to record the time it takes to hit the ground. if it takes 7.0 s for the rock to return to the same location from which it was released, what is the acceleration due to gravity on the planet?

Answer 1

Interesting ...

gravity can be considered as *constant* acceleration, then you have:

h = v0*t + 1/2*g*t^2,

Final height is the same as the initial = 0

0 = 11*7 +1/2*g*7^2 ---> g = -11*7*2/7^2 = -22/7 m/s^2

Minus because it goes down, You can say g_planet = 22/7 m/s^2 ~ 3.14 m/s^2. Earth's one on its surface is about 9.81 m/s^2, so this planet has 3 times less gravity. It may be 1/3 the radius of the Earth, assuming same density. This is not asked! :)

Answer 2

Answer: g = -3.143 m/s²

Explanation: To determine acceleration due to the gravitational force, it can be used the following formula:

v = v₀ + gt, where:

v₀ is the initial velocity;

g is acceleration due to gravity

t is the time to return to the point of origin

When the rock return to the point of origin, there is no velocity, so v = 0.

The time to go up is the same to go down, so t = $\frac{t}{2}$ = $\frac{7}{2}$.

Substituing in the formula:

v = v₀ + gt

0 = 11 + g.$\frac{7}{2}$

g = $\frac{(0 - 11).2}{7}$

g = - 3.143

The acceleration due to the gravitational force is g = - 3.143 m/s².