Question

An object is falling freely from rest near the surface of the earth. It takes `2"s"` for the object to reach the ground `64"ft"` below. The acceleration of the object must be equal to `-g`: true or false?

Answer 1

True.

Explanation:

I believe the purpose of this question is to get you to think about what the acceleration of an object in free-fall should be, for which the situation meets the conditions to apply reasonable assumptions (e.g. air resistance is negligible).

We are told that the object is falling freely from rest near the surface of the earth, which means there is a potential for air resistance to not play such a big role that we cannot confidently state that the acceleration of the object should be about `9.81"m"//"s"^2 (~~32"ft"//"s"^2)`. However, we are not told what the object is! If the object is a piece of paper, for example, we cannot make this statement. Let's calculate the acceleration and see what we get before we discuss further.

We are given that `Deltat=2"s"` and `Deltay=64"ft"=19.5072"m"`. I have converted to SI units for more simplicity in the comparison of our accelerations. Assuming a constant acceleration, we can use the kinematic equation:

`y_f=y_i+v_iDeltat+1/2a(Deltat)^2`

Since the object begins from rest, we know that `v_i=0` and we can assume that the final position, i.e. the ground, is located at `y_f=0`.

`=>0=y_i+1/2a(Deltat)^2`

Solving for the acceleration `a`:

`=>a=(-2y_i)/(Deltat)^2`

Using our known values

`=>a=(-2(19.5072"m"))/(2"s")^2`

`=>=-9.7536"m"//"s"^2`

`=>~~-9.8" m"//"s"^2`

Which is the free-fall acceleration, `-g`.

This is to be expected for objects of reasonable mass/surface area (i.e. for objects/situations for which factors like air resistance are said to be negligible).

So, it looks like your answer is true.