An object falls freely from rest on a planet where the acceleration due to gravity is it is on earth. in the first 5 seconds it falls a distance of twice as much as

a) 250 m.
b) 500 m.
c) 150 m.
d) 100 m. e) none of these

A. The object falls a distance of 250 m

Explanation:

Hi there!

In the question, you have forgotten the acceleration due to gravity. However, looking on the web I´ve found a very similar problem in which the acceleration due to gravity was as twice as much as it is on Earth.

The equation of height of a falling object is the following:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the object after a time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (on Earth: ≅ -10 m/s² considering the upward direction as positive).

Let´s place the origin of the system of reference at the point where the object is released so that y0 = 0. Since the object falls from rest, v0 = 0.

Then, the height of the object after 5 s will be :

y = 1/2 · 2 · g · t²    (notice that the acceleration due to gravity is 2 · g)

y = g · t²

y = -10 m/s² · (5 s)²

y = -250 m

The object falls a distance of 250 m.

the moving of the box is a result of centripetal force. think of that  as a car or a truck in this case creates a turn, the force of friction performing upon the turned wheels of the car runs centripetal force obligatory for circular motion for the things inside it.