# 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:

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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. $v_{avg}=\frac{422}{3.4905 + 19.248}= 18.56ms^{-1}$