Calculating Intitial Position given the ratio between the last two seconds of free fall for 1D motion

Question

I've tried this problem in multiple different ways and can't seem to come up with an acceptable answer. The question is,

A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 48 % of the entire distance it falls.

I tried using kinematic equations to resolve this in multiple different ways. For example, setting the position function at t-1 equal to .48 times the position function of t, and then solving for t. I then plugged in t into the position function, and got 52m, which is not correct.

Our textbook was kind enough to give a similar problem, except it used 45% and got the height to be equal to 73m. What is the proper way to go about solving this problem? What am I doing wrong?

Answer 1

This necro-bumped question deserves an answer that touches the delicate matter of finding the right solution when many roots exist.

In most cases, the "wrong" solutions can still be understood what they would hypothetically mean, even if they have no physical realization.

Let's say the time gap (last second) is $t_0$ and the displacement ratio to the total height is $\eta =z/h$ (to avoid explicit numbers).

We have

$$h=\frac12 gt^2;\quad \eta h=\frac12 g(t-t_0)^2$$

Subtracting these, we obtain $$h(1-\eta)=\frac12 g(2tt_0-t_0^2)$$

Plug it back into the first equation:

$$\frac12 g(2tt_0-t_0^2)\frac{1}{1-\eta}=\frac12 gt^2$$ Cancel and rearrange: $$(1-\eta)t^2-2tt_0+t_0^2=0$$ Solve: $$t=t_0\frac{1\pm \sqrt\eta}{1-\eta}=t_0\frac{1}{1\mp\sqrt{\eta}}$$

Your numbers $\eta=0.48$ and $t_0=1\,\rm s$ give solutions $t=\{3.3{\,\rm s},0.6{\,\rm s}\}$

Here, we have to discard the second solution, as it would measure the height between times 0.6s and -0.4s, which was before the fall started (where it would have been, if it wasn't just dropped from the top but thrown upwards, and we just started the clock at the highest position).

If we plug both solutions into the first height equation $\frac12 gt^2$, we get

$$\color{green}{ h(3.3{\,\rm s})=52\,\rm m},\quad \color{red}{h(0.6{\,\rm s})=1.7\,\rm m}$$

Notice how the first solution (computed by the original poster) is correct, as the second one corresponds to the one where the travel time is shorter than the measurement interval.

Answer 2

Given the standard kinematics equation $$h = \frac{1}{2} g t^2$$ you have two unknowns, the time $t$ and height $h$.

But with the additional piece of information you can create another equation $$ 0.48 h = \frac{1}{2} (t-1)^2$$ for the same two unknowns.

So two non-linear equations and two unknowns. You subtract one from the other

$$ h - 0.48 h = \frac{1}{2} g ( t^2 - (t-1)^2 ) $$ to results in a linear relationship between $h$ and $t$ and then use one of the original equations to solve the problem. But the solution will have two roots. It makes sense to pick the one with the least time of the two, and that is not the one with 52m as the answer.