Basic Motion Problem (Dropping Objects)

[doug1]

Homework Statement

You are given a length of rope, tape and five rocks. The rocks must be attached in such a manner that when the rope is released, the sound of each rock hitting the ground is evenly spaced.

There must be a rock at the very bottom and the very top of the rope. When the rope is released, the bottom rock will be just above the ground. A stopwatch cannot be used (time cannot be recorded).

Homework Equations

Any of the following kinematics equations:

d=1/2(Vf+Vi)t
Vf=Vi+at
d=Vit+1/2at^2
d=Vft-1/2at^2
Vf^2=Vi^2+2ad

The Attempt at a Solution

a.) I understand that the rope will accelerate downwards due to gravity. This means that the rock at the top of the rope will have a faster final velocity just before impact than the rock at the bottom of the rope.
b.) Assuming that the bottom rock takes 0.1s to hit the ground I made the following calculations:

(Trying to space the sounds 0.2 second apart):

2nd rock from the bottom:

d = Vit+1/2at^2
d = (0)t + 1/2(9.81m/s^2)(0.3s)^2
d = 0.44 m

This means the second rock will be placed 0.44 m from the first rock (0.44 m from the bottom of the rope).

3rd rock from the bottom:

d = Vit+1/2at^2
d = (0)t + 1/2(9.81m/s^2)(0.5s)^2
d = 1.23 m

This means that the third rock will be placed 1.23 m from the bottom of the rope.


Is this the correct approach?

 

[CWatters]

Close but you are given the length of rope so the distances have to add up to that length. Hard to do it that way.

Perhaps write equations for d_top and d_bot in terms of Δt. where Δt = one time interval.

The length of the rope is then d_top - d_bot. Rearrange to give Δt ?

Something like that.