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

1 Answer
Aug 15, 2016

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.