Gravity, height, and rock falling problem - Please help ASAP

Question

A rock dropped from a cliff covers one-third of its total distance to the ground in the last second of its fall. Air resistance is negligible. How high is the cliff?

Answer 1

Hi Ben: I can see why you might need help with this one. It’s an elegantly stated little problem which looks deceptively simple, until you start trying to solve it.

There are a couple of things you might recognize before you even put pencil to paper. First, we are tempted to say the distance the rock falls in the last second is D = ½at². That’s certainly a familiar formula but it won’t work here. Why? Because the rock is already falling when the last second starts.

D = ½at² gives us the distance a rock falls when it starts from a dead stop, not when it’s already hurtling downward when the last second starts. Watch for this trick in story problems where birds or planes suddenly swoop downward. They have a downward velocity when they start. Another way this problem can be presented is if you are given the speed of say a textbook as it passes the third floor window and you are told to calculate the distance or time to the ground. You need a formula more like

D = V₀t + ½at2. The V₀ is the initial speed downward. In some cases (like this problem) you can even find a way to avoid working with velocity at all.

The next problem here is that you need to know how fast the rock is traveling downward when it starts the last third of the journey, but in order to know that, you need to know how far or how long it has already fallen, but in order to know that you need to know how fast it is moving, but in order to know that . . . you see where I’m not going with this.

Whenever you see this kind of pattern, you are probably going to have to find two equations and solve them simultaneously. The simplest way to do that is usually to solve for a variable in one equation, then substitute it into the other equation. For this problem let’s consider these variables and equations:

Let D = the total distance the rock falls and the final answer to the problem.

Let D₁ = the distance of the first leg of the journey, up until time = t-1 second, 2/3 of the total.

Let D₂ = the distance of the second leg of the journey, from t-1 to t, the last third.

Then we can consider these equations: 2/3 D = D₁

1/3 D = D₂

D = D₁ + D₂

You can use any of them but I think the easiest and shortest one is 2/3D=D₁. This was not immediately obvious to me, so I did them all, but I will do the easiest one here.

D = V₀t + ½at² Memorize this equation if you haven’t already. In this problem we are using it to say the total distance traveled is the initial velocity (0) times the total time traveled, plus ½ the acceleration due to gravity times the total time squared. So:

D = (0)t + ½ at² = ½ at²

D₁ = V₀t + ½ a(t-1)² It’s the same equation, just modified for the first 2/3 of the journey (t-1).

D₁: = (0)t + ½ a(t-1)² = ½ a(t²-2t+1) = ½ at² – at + ½ a

Now we can substitute values for D and D₁ into 2/3 D = D1 giving:

2/3 * ½ at² = ½ at² – at + ½ a

1/3 at² = ½ at² – at + ½ a

2/6 at² = 3/6 at² – at + ½ a

0 = 1/6 at² – at + ½ a

Now if you apply the quadratic formula (ignoring the value of "a" because you can divide out a constant without changing the value of the roots) you will see that t = 3 ± v6.

3 - v6 can be discarded because although it is a root of the equation, it does not meet the constraints placed on us by the story problem. Namely, the trip is too short. So the total time traveled is 3 + v6.

Now we substitute back into our equation for D. I will use 9.8 m/s2 for the acceleration of gravity, giving us our final answer in meters:

D = ½ at² = ½(9.8)(3+v6)² = 145.51 meters

Check: D1 = ½ at² – at + ½ a

               = ½(9.8)(3+v6)² – (9.8)(3+v6) + ½(9.8) = 97.01 which is 2/3 of D

Answer 2

We can use the formula D = (1/2)(acceleration)(time)². Using this, we can determine the distance covered in the last second:

d=(1/2)(9.8m/s²)(1 sec)² = 4.9 m

We know that 1/3 the distance is covered in the last second, so the cliff is 3*4.9m = 14.7 m