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

In summary, the parachutist loses a shoe at an altitude of 20 meters and loses velocity as they fall. The shoe reaches the ground in 1.16 seconds. The final velocity is 10.3 meters per second.
  • #1
mortho
100
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
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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!
 
  • #6
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?
 
  • #7
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.
 
Last edited:
  • #8
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!
 
  • #9
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.
 

1. What is the definition of free fall in regards to a parachutist?

Free fall is the motion of an object that is only acted upon by the force of gravity. In the case of a parachutist, this means that they are falling towards the ground without any external forces acting on them.

2. How does air resistance affect a parachutist in free fall?

Air resistance, also known as drag, is a force that opposes the motion of an object through air. As the parachutist falls through the air, they experience a force of air resistance that increases as their speed increases. This eventually balances out the force of gravity, causing the parachutist to reach a terminal velocity and fall at a constant speed.

3. What factors influence the rate at which a parachutist falls?

The rate at which a parachutist falls is influenced by several factors, including the mass and shape of the parachutist, the surface area of the parachute, the altitude at which the jump takes place, and the presence of any external forces such as wind.

4. How does the size of a parachute affect free fall for a parachutist?

The size of a parachute affects free fall for a parachutist by increasing the amount of air resistance they experience. A larger parachute will have a greater surface area and thus create more drag, slowing down the parachutist's descent. This allows for a slower and safer descent to the ground.

5. Is free fall for a parachutist a constant acceleration?

No, free fall for a parachutist is not a constant acceleration. While the parachutist is initially accelerating due to the force of gravity, as they reach terminal velocity, their acceleration decreases and becomes constant. This is because the force of air resistance is equal to the force of gravity, resulting in a balanced net force and constant velocity for the parachutist.

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