While I was studying about finite differences I noticed a question in difference equations.Does anyone knows how to solve this using difference equations?
WORD PROBLEM
Imagine you are to jump from an aircraft at an altitude of 1000 metres. You want to free-fall for 600 metres, knowing that in successive seconds you fall 5, 15, 25, 35, ... metres. How many seconds do you count before you pull the rip-cord?
Let $d_n$ be the total distance fallen after $n$ seconds. Looking at the pattern, we see that $d_1 = 5$, $d_2 = 20 = 5 + 15 = d_1 + (10 \cdot 2 - 5)$, $d_3 = 45 = 20 + 25 = d_2 + (10 \cdot 3 - 5)$, and for all $n > 1$: $$ d_n = d_{n-1} + (10n - 5) $$ Let's see what happens as we replace $n$ with $n-1$ successively: \begin{align*} d_n - d_{n-1} &= 10n - 5 \\ d_{n-1} - d_{n-2} &= 10(n - 1) - 5 \\ d_{n-2} - d_{n-3} &= 10(n - 2) - 5 \\ d_{n-3} - d_{n-4} &= 10(n - 3) - 5 \\ &~~\vdots \\ d_{2} - d_{1} &= 10(2) - 5 \\ \end{align*} Summing the $n - 1$ equations together, notice that the LHS telescopes, yielding: $$ d_n - d_1 = \sum_{k=2}^n(10k - 5) $$ The RHS is just an arithmetic series of $N = n - 1$ terms with initial term $a = 15$ and last term $L = 10n - 5$, so we obtain: \begin{align*} d_n - 5 &= \frac{n - 1}{2}(15 + (10n - 5)) \\ d_n &= \frac{n - 1}{2}(10(n + 1)) + 5 \\ d_n &= 5(n^2 - 1) + 5 \\ d_n &= 5n^2 \\ \end{align*} Letting $d_n = 600$, we conclude that the rip-cord should be pulled at $n = \sqrt{\frac{600}{5}} = 2\sqrt{30} \approx 10.95$ seconds.