# Free fall parachutist problem

## Homework Statement

A parachutist descending at a speed of 20.0 m/s loses a shoe at an altitude of 20.0 m. (Assume the positive direction is upward.) a)When does the shoe reach the ground? (seconds) b)What is the velocity of the shoe just before it hits the ground?

## Homework Equations

i used Vf=square root(vi2 + 2ax) or Vf=squareroot(20squared+2*20*-9.81)
for t i used t =Vav/a but i don't think that's a correct equation

## The Attempt at a Solution

for Vf, i got 2.76 m/s and for time i got 1.16 s
But i know the time is definitely wrong because it has to be less than 1 second since both speed and distance are 20.0

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Dear God, please don't let the parachutist hit the ground at 20 m/s.

Do you know how to do quadratic equations? The standard equation that solves all such problems is
X = Xo + Vot + 0.5at^2 but you must be able to do quadratics.
If you use the "canned" equations like
Vf=squareroot(20squared+2*20*-9.81)
you need to be able to account accurately for directions. Go back and look at directions, that is, what is up and what is down.

no i have just started physics so i have not learned quadratics yet. I've been trying to solve it for the past couple of minutes and i tried t=squareroot(2a/x) so i did t=squareroot(2*-9.81/20.0) and i got 0.99 seconds which looks right. Then for the Vf i used Vf=Vi+at so Vf=20.0+(-9.81*0.99) and got 10.3 m/s . Is that correct?

initial velocity (v0) = 20 m/s
gravity (g) = ~10 m/s/s
distance (s) = 20 m

(a) what is time (t) = ?
(b) what is the final velocity (vf) = ?

Use your "big five" formula for accerated motion:
(a) s= (v0)(t) + (1/2)(g)(t^2)

You know g, v0, and s, so plug them in and solve for t.

Once you know time, you can easily find vf with another big five forumula:
(b) vf= v0 + at

Familiarize yourself with these formulas, I use them all the time for physics--they're very important.

(a) t= 0.8 s
(b) vf= 28 m/s
if I did not make any computational mistakes.

If you use 9.8, or 9.81 for gravity, your answers will be more precise, which the significant figures require for this problem.

no i have just started physics so i have not learned quadratics yet. I've been trying to solve it for the past couple of minutes and i tried t=squareroot(2a/x) so i did t=squareroot(2*-9.81/20.0) and i got 0.99 seconds which looks right. Then for the Vf i used Vf=Vi+at so Vf=20.0+(-9.81*0.99) and got 10.3 m/s . Is that correct?
By the way, "quadratics" is algebra, not physics!

okay so i get how you got the Vf but for the time you used gravity? is that supposed to be the same as the -9.81m/s2? how did you set up the formula for the time?

Yeah, like I said, depending on the precision you want you'll use more significant figures for gravity.

The formula you use to find time: (where 's' is distance)
s= (v0)(t) + (1/2)(g)(t^2)

Plug in your knowns, and solve for t.
This is a formula you should memorize for acceleration problems. It is derived from the "Mean Speed Theorm" and definition of acceleration. Our teacher made us memorize this equation and four others for kinematics/acceleration.. you should too depending on the physics course you are taking. If you are taking a college level course you definitely should memorize them. They're call the "Big Five" equations.

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okay so for your answer you had t=0.8 s so how did you get that? because i tried all these different ways and i kept getting different numbers. could you show me the equation you used. i know you used the big five one because i also used that but i don't know if i put it in the right way for time. Thanks!

Sorry I took so long:
s= (v0)(t) + (1/2)(g)(t^2)

20=(20)(t) + (1/2)(9.81)(t^2)
so 0= 20t + 4.905t^2 - 20
arrange it: 0= 4.905t^2 +20t - 20

Now solve for time using algebra, the quadratic formula.
I double checked, and using 9.81 instead of 10 for gravity I got the answers 0.8307 and -4.908. Since time cannot be negative, the answer is 0.8307 s. If you have trouble getting from "0= 4.905t^2 +20t - 20" to 0.8307, you should review your math. Your physics looks good though if you can get that far.