# Homework: A parachutist jumps from a plane [closed]

I need help with the following problem. I think I almost have it . . .

A parachutist bails out and freely falls a distance of $y_1$. Then the parachute opens and thereafter, the parachutist decelerates at a rate of $a_2$. She reaches the ground with a speed of $v_2$. Find her average speed for the fall. (The answer should be $18.5\ \text{m/s}$.)

Note: In the diagram, replace "x" with "y". :) $\text{Let upwards be the positive y direction.}\\ \text{Givens:}\\ y_1 = -59.7\ \text{m}\\ v_0 = 0\ \text{m/s}\\ v_2 = -3.41\ \text{m/s}\\ a_1 = -9.8\ \text{m/s$^2$}\\ a_2 = 1.60\ \text{m/s$^2$}\\ \text{Solve for$t_1$:}\\ y_1 - y_0 = \frac{1}{2}a_1t^2+v_0t\qquad\text{for}\ \ y_1 = 0; v_0 = 0; t=t_1.\\ t_1 = \pm\sqrt{\frac{2y_1}{a_1}}\\ t_1 \approx \pm 3.4905\ \text{s}.\\ \text{Plug$t_1$into the velocity equation for$v_1$to find$v_1$:}\\ v_1 = a_1t_1 + v_0\\ v_1 \approx -34.207.\\ \text{Plug$v_1$into the velocity equation for$v_2$to find$t_2$:}\\ v_2 = a_2t_2 + v_1\\ t_2 = \frac{v_2 - v_1}{a_2}\\ t_2 \approx 19.248\ \text{s}.\\ \text{Plug$t_2$into the position equation for$y_2$to find$y_2$.}\\ y_2 = \frac{1}{2}a_2t_2^2+v_1t_2 + x_1\\ y_2 \approx -422\ \text{m}.\\ \text{Now solve for$\lvert\overline{v}\rvert$:}\\ \lvert\overline{v}\rvert = \lvert\frac{x_2 - x_0}{t_2 - t_0}\rvert\\ \lvert\overline{v}\rvert \approx 21.9\ \text{m/s} \neq 18.5\ \text{m/s}.$

My solution is incorrect. What am I doing wrong?

## closed as off-topic by John Rennie, Abhimanyu Pallavi Sudhir, Kyle Kanos, Brandon Enright, Waffle's Crazy PeanutJan 25 '14 at 14:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Abhimanyu Pallavi Sudhir, Kyle Kanos, Brandon Enright, Waffle's Crazy Peanut
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1. The value you calculated for $t_2$ is the time taken for the person to fall the distance $y_2-y_1$
2. Thus, the total time to fall the total distance of $422m$: $$T= t_1 + t_2$$ $$=3.4905 + 19.248$$
4. $v_{avg}=\frac{422}{3.4905 + 19.248}= 18.56ms^{-1}$