Question

Could someone please explain in detail the effect of height on the time an object takes to fall?

When a paper "robocopter" is dropped from heights that increase by 0.5m every time, the time it takes to fall is longer every time. Is this because the air resistance is more or just because there is more air to fall through? Thank you!!

Answer 1

It takes longer to fall because of the increasing height.

Explanation:

The air resistance will be the same as long as the shape of the robocopter stays the same, so that would not be a factor. The increase in displacement (height/distance) each time is the reason that the time to fall increases with each trial.

The following equation can illustrate this.

`d=v_it+1/2at^2`, where `d`=displacement (distance/height), `v_i`=initial velocity, `a`=acceleration, and `t`=time.

Given/Known
`d=-"0.500 m"`
`v_i=0`
`a=g=-"9.81 m/s"^2"`
`g` is acceleration due to gravity
Because the motion is downward, both the displacement and acceleration are negative.

Unknown
time, `t`

Rearrange the equation to solve for `t`.

Since `v_it=0*t=0`, we can omit it from the equation.

`d=1/2g*t^2`

Multiply both sides by `2`.

`2d=g*t^2`

Divide both sides by `g`.

`(2d)/g=t^2`

Switch sides.

`t^2=(2d)/g`

Plug in the known variables.

`t^2=(2xx-0.500"m")/(-9.81 "m/s"^2")`

Take the square root of both sides.

`sqrt(t^2)=sqrt((2xx-0.500"m")/(-9.81 "m/s"^2"))`

`t="0.319 s"`

Since both the displacement and acceleration are negative, they cancel each other when divided, so the result is positive.

Redo the equation when the displacement (height) is `-"1.00 m"`

`sqrt(t^2)=sqrt((2xx-1.00"m")/(-9.81 "m/s"^2"))`

`t="0.452 s"`

Now redo the equation with the displacement of `-"1.50 m"`.

`sqrt(t^2)=sqrt((2xx-1.50"m")/(-9.81"m/s"^2")`

`t="0.553 s"`

As you can see, as the height displacement (height) increases, the longer it takes for the robocopter to fall.