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# Free fall. Distance passed task.

Solution of the falling body and distance passed tasks.

There is the following formulation of the problem in the request /4070/ made by the user: Free falling body without initial velocity passed 2/3 of its path at last second of its fall. Find the entire distance traveled by the body.

The same task can be formulated more simple: A free falling body with initial velocity v0 passed K part of its distance for the last n seconds. Find the entire distance the body traveled.

It's possible to create a calculator for such formulation. Actually, just below you will find a calculator, solving the task for given v0, n, K and g (free fall speed). And the process of solving is written in formulas below the calculator.

### Free fall. Distance passed task.

Digits after the decimal point: 2
Fall time

Distance passed

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So, how to solve the problem. To begin with, we write the known conditions as formulas:
Total distance the body passed
$S=V_0t+\frac{gt^2}{2}$

Distance passed without the last n seconds of the fall
$S'=V_0(t-n)+\frac{g{(t-n)}^2}{2}$

It is known that the difference between the two is a certain amount of K of the total distance
$V_0t+\frac{gt^2}{2}-V_0(t-n)-\frac{g{(t-n)}^2}{2}=K(V_0t+\frac{gt^2}{2})$

Next, expand the brackets, translate everything left, reduce and sort the monomials by degree t. As a result, we obtain the following
$-\frac{Kg}{2}t^2+(gn-KV_0)t-\frac{gn^2}{2}+V_0n=0$

We get the conventional quadratic equation with respect to t with the following coefficients
$a=-\frac{Kg}{2}$$b=gn-KV_0$$c=-\frac{gn^2}{2}+V_0n$

All the figures for the calculation of the coefficients are known from the condition. And I hope everybody can solve quadratic equations.

We have to choose the correct root only. It's obvious, that if there are no roots, there is no solution. If there are, then the desired root must be greater than n (otherwise it makes no sense), and if both of the roots more than n, the smaller one will suit us.

When we find the time, the distance passed can be found by substituting the first formula.
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