When does the shoe reach the ground and what is its velocity?

  • Thread starter Thread starter mortho
  • Start date Start date
  • Tags Tags
    Fall Free fall
Click For Summary

Homework Help Overview

The problem involves a parachutist who loses a shoe while descending, requiring the determination of the time it takes for the shoe to reach the ground and its velocity just before impact. The context is kinematics, specifically dealing with motion under gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various equations of motion, including the use of kinematic formulas and the quadratic equation. There are attempts to calculate time and final velocity using different methods, with some questioning the accuracy of their approaches and assumptions about direction.

Discussion Status

Multiple interpretations of the problem are being explored, with participants providing different methods and results for time and velocity. Some guidance on using specific kinematic equations has been shared, and there is an ongoing dialogue about the correct application of these formulas.

Contextual Notes

Some participants express uncertainty about the use of gravity's value and its direction, while others note the importance of significant figures in their calculations. There is a mention of homework constraints regarding the use of certain mathematical techniques, such as quadratics.

mortho
Messages
99
Reaction score
0

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
 
Physics news on Phys.org
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 + volt + 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.

The answers I got were
(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.
 
mortho said:
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 theorem" 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.
 
Last edited:
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.
 

Similar threads

  • · Replies 34 ·
2
Replies
34
Views
4K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
11
Views
3K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 3 ·
Replies
3
Views
1K