# Skydiver jumping from a plane

1. Dec 25, 2014

### Barclay

1. The problem statement, all variables and given/known data

Q 1. How fast will a skydiver be travelling 1 second after jumping from a plane?

2. Relevant equations

SUVAT formula

3. The attempt at a solution

I guessed the answer to be 10 m/s because something to do with gravity and a 1 kg mass. The answer is correct according to the book but I’m not sure why. The mass of the skydiver will be much more than 1 kg. So my guess is not really an educated guess.

Please can you explain to me the answer in words terms (without the use of equations). (If possible) Thank you.

I recently learnt about the SUVAT formula from another post recently so have applied this here. (Please note that the book I’m reading does not even mention SUVAT formula yet so I don't think I meant to be using SUVAT)

s = distance
u = initial velocity
v = final velocity
a = acceleration
t = time

v = u + at

V = 0 + 10*1 (I am assuming acceleration to be 10m/s2 because of gravity)

V = 10m/s (The answer is correct according to the book)

I’d like to post the second part of this question here too, if I may because it is linked to the first question.

Q 2. If the air resistance force on the skydiver is 0.1V2, what will be its magnitude and direction after one second?

Here I have no idea where to start. Please note this section of the book is discussing terminal velocity (if that has any relevance).

Answer in the book is 15 N upwards

2. Dec 25, 2014

### SteamKing

Staff Emeritus
The velocity at which a body falls is independent of its mass, at least in a vacuum. In an atmosphere, drag tends to slow falling objects, but for such a short interval, namely 1 second, the amount by which the velocity is reduced can be neglected.

The acceleration due to gravity, approximately 10 m/s2, means that the velocity of a falling object will change approximately by 10 m/s for each second the body falls. So, if a body is dropped from a great height, after one second, the body is falling at 10 m/s, after 2 seconds, it is falling at 20 m/s, etc., etc.

3. Dec 25, 2014

### SteamKing

Staff Emeritus
What is the velocity of the skydiver after 1 second? [Hint: see answer to Q1.]

Use this result to calculate the magnitude of the drag, according to the formula. What is the direction of the drag force on a falling body?

4. Dec 26, 2014

the answer to the question is quite simple.the acceleration due to gravity does not depend on the mass of the object(g=GM/R^2;M is the mass of earth).at all points,the acceleration will 9.80665 m/s^2, HOWEVER,if the aircraft is flying at a sufficient height,then the net force will be zero(hence zero acceleration) as it will be moving at terminal velocity,which is-v=√2mg/ρACd.
For Q2,first find out the velocity(you know that the acceleration is 9.8m/s^2≈10m/s^2 and time is 1 second) and use the formula to calculate the magnitude of drag.
Try to think,in which direction would air resistance(drag) act,opposite or in the direction of motion?

5. Dec 26, 2014

### CWatters

The definition of acceleration is the change in velocity divided by the change in time eg...

a = (v-u)/t

Rearange that and you get the change in velocity (v-u) = at.

Rearrange that and you get

v = u + at

In this problem u=0 so

v = at = 10 * 1 = 10m/s

6. Dec 26, 2014

### Barclay

Q 2. If the air resistance force on the skydiver is
0.15V2, what will be its magnitude and direction after one second?

I made an error above. I was meant to write 0.15V2 but accidentally wrote 0.1V2

I apologise for any inconvenience caused

Last edited: Dec 26, 2014
7. Dec 26, 2014

### Barclay

Have been trying the calculation according to Steam King's suggestion but not succeeded.

There's no mention that there is a terminal velocity after 1 second so the upwards (air resistance) force and downwards gravity cannot assumed to be equal.

Anyway have tried using F= ma but there's no mention of mass of the skydiver (that comes in the next question in the book).

Tried SUVAT equations but don't work either. No idea

8. Dec 26, 2014

### Barclay

Just figured it out .... after few hours : AIR RESISTANCE = 0.15V2

We know V = 10m/s so 0.15 * 102 = 15

Answer = 15 N upwards force

Thank you Steam King and everyone else. Such an obvious solution that duped me for hours then came to me in a flash after looking at Stem King's suggestion for the ump-teenth time:(

Last edited: Dec 26, 2014
9. Dec 26, 2014

### rcgldr

The question should have been worded so that the skydiver is jumping off some stationary object. If jumping from a plane, the skydiver starts off with a horizontal velocity well in excess of 100 kph, and the aerodynamic drag is already significant. Ignoring any "lift" issues related to horizontal velocity, the vertical component of velocity would start at 0 kph and increase as shown in the math in the previous posts.

10. Dec 27, 2014

### haruspex

I agree with your concern about the wording, but wouldn't that huge horizontal airflow greatly reduce the downward acceleration, regardless of any lift? Or are you using 'lift' in a somewhat loose manner?

11. Dec 27, 2014

### rcgldr

Part of the issue is that at the start of the jump from a plane, the skydiver could be oriented to create a positive, zero, or negative angle of attack. Even with a positive angle of attack, the skydiver would still accelerate downwards (assuming the skydiver isn't using a high lift wing suit), but at a slower rate than just a vertical drop.

I've also read some projectile with aerodynamic drag articles that state that the x (horizontal) and y (vertical) components of drag are related to

$v_x v$ and $v_y v$ and not $v_x^2$ and $v_y^2$

but this would seem to conflict with the common example that a dropped bullet hits the ground at the same time as a fired bullet (assuming a flat earth).

Last edited: Dec 27, 2014
12. Dec 27, 2014

### haruspex

Yes, this is my point. Treat the diver as a sphere. If the x and y velocity components are vx, vy then the drag has magnitude k(vx2+ vy 2). The y component of this is $k v_y\sqrt{{v_x}^2+{v_y}^2} = k v_y v$.
At first, the x component of velocity is much the greater, so the y component approximates $k v_y v_x$. Of course, vx will diminish quite quickly, but while it is large the vertical drag can be rather greater than given by $k {v_y}^2$.
I modelled the question in 0.05 second steps, g = 10 m/s, and a diver mass of 70kg. With no horizontal speed, the vertical drag acting over the second cut the final speed from 10 m/s to 9.93m/s. With an initial horizontal speed of 30m/s, the vertical speed after 1 sec was 9.70 m/s.