Alaskan Rescue Plane

Warmacblu

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

An Alaskan rescue plane traveling 40 m/s drops a package of emergency rations from a height of 113 m to a stranded party of explorers.

G = 9.8 m/s^2

Where does the package strike the ground relative to the point directly below where is was released? Answer in units of m.

What is the horizontal component of the velocity just before it hits? Answer in units of m/s.

What is the vertical component of the velocity just before it hits? (Choose upwards as the positive vertical direction) Answer in units of m/s.

2. Relevant equations

Square root of 2h/g = time of free fall

y = Voy(t) - 1/2gt^2

x = Vox(t) + Xo

3. The attempt at a solution

I solved the first part using 40 m/s x (square root of 2h/g) = 192.09 m

The answer to the second part was 40 because that is the constant horizontal component.

For the third part, I tried solving for T using x = Vox(t) + Xo and plugging that into y = Voy(t) - 1/2gt^2 but could not come up with the correct answer. I believe I need to include the time it takes to hit the ground somehow but do not know.

Any help is appreciated.

Thanks,
Ben

Chewy0087

part 1 & 2 look okay to me

for part 3)
hm for a moment ignore the horizontal component, only the vertical, well you know how far it was dropped from - displacement - 113m, you know it's acceleration - g, and you know its initial speed when it was dropped from the plane - 0, because it was still, now use your equation to find the time taken to reach the ground;

y = Voy(t) - 1/2gt^2

kuruman

Use

v_y=v_0y - g t

You know everything except v_y.

Warmacblu

I tried this:

Vy = Voy - gt
Vy = 47.06 - (-9.8 x 4.80)
Vy = 94.1 which was an incorrect answer

If I use a positive 9.8 value, I get .02 which seems awfully low.

mgb_phys

What is the initial vertical velocity of the dropped package ?

Warmacblu

47.06

I used y = Voy(t) - 1/2gt^2

113 = Voy(4.80) - 112.896
225.896 = Voy(4.80)
Voy = 47.06

I thought this was the correct answer because the vertical component of the velocity should be the initial velocity in the y-direction, no?

mgb_phys

If you drop something isn't the initial vertical velocity zero ?

Warmacblu

Yes, but the vertical component of the velocity would change and would not be zero just before it hits.

Should I alter my y-component equation to something like:

Voy = -1/2gt^2

mgb_phys

In your equation Voy is the initial vertical velocity.
If you drop something the initial velocity must add to the acceleration, you have the opposite sign as if velocity was in the opposite direction to the acceleration

Warmacblu

So going back to my first equation:

y = Voy(t) - 1/2gt^2

Am I supposed to use -9.8 for my g value or positive 9.8. If I use - 9.8, the math works out like this:

y = Voy(t) - 1/2gt^2
113 = Voy(4.80) + 112.896
.104 = Voy(4.80)
Voy = .0216

To me this makes more sense because I would think that a package should arrive slowly so that the contents aren't ruined. A package traveling at 47.06 m/s would be destroyed upon impact.

kuruman

How did the negative sign in front of (1/2)g t² become positive when you put in the numbers? You do not seem to understand the meaning of

v_y=v_0y - g t

It says that for every second that goes by, you add -9.8 m/s to the vertical component of the velocity. The negative sign means in the "down" direction. If the package is in the air for 4.8 s you need to add 9.8*4.8 m/s in the down direction to the 0 m/s that is the initial vertical velocity. Yes. that makes a lot of meters per second, but that's why people use parachutes to drop packages (or themselves) from planes.

Warmacblu

Truthfully I always seemed to get confused on when to use a negative sign in front of 9.8. The negative sign became positive because it was minus negative 112.896. I thought that a double negative equals a positive. Or was I supposed to use a positive 9.8 value?

Better explanation on what I did:

-1/2gt^2

(-1/2)(-9.8)(4.80^2) = 112.896

Jebus_Chris

[tex]\Delta y = v_ot - \frac{1}{2}gt^2[/tex]
[tex]\Delta y = v_ot + \frac{1}{2}gt^2[/tex]
You can write the equation either way. It's all about sign convention. Generally when you write the first form you take the negative sign from g, that way you know that the object would be falling.
How you determine the sign of g (or any a). First you have to determine a sign convention, that is which way is positive velocity and which way is negative velocity. For example, dropping a ball from a height. You can make g negative and the h negative, or you can just make them both positive so you don't have to deal with negatives(you just have to realize it's really falling down). If you throw a ball downward from a height you can make all 3 negative or positive, again it's easier with positives. If you throw a ball up then it's initial velocity will be positive and g will be negative.
If acceleration is in the same direction as motion, then it should have the same sign as the velocity. If acceleration is in the opposite direction of motion, it should have the opposite sign of velocity.

Warmacblu

Ahh, okay. So it's similar to drawing a picture where you want to set your point of reference. Thanks for the explanation. As for solving the problem:

V_y = V_0y - g t

V_y = 0 - (9.8 x 4.8) = -47.04 which was correct.

My last question regarding this topic is how did you derive that equation or is it just a handy one to keep around?

Jebus_Chris

Derived what equation?

kuruman

The derivation is simple. "Acceleration" means "change of velocity with respect to time." When the acceleration is constant the rate of change of velocity is constant. What does that mean? That in equal time intervals, the velocity changes by equal amounts. So if the acceleration is -10 m/s/s and the velocity is 40 m/s at time zero, then it will be

40 - 10*1 m/s at time t = 1s
40 - 10*2 m/s at time t = 2s
40 - 10*3 m/s at time t = 3s
etc.

This works equally well with any initial velocity v_0y, not just 40 m/s, and any acceleration a not just -10 m/s/s. So in general

v_y = v_0y + a*t

Note that in this case we have defined "up" as positive and "down" as negative. Therefore, a = - 9.8 m/s/s, where the negative sign indicates that the acceleration is down. The symbol "g" stands for the magnitude of the acceleration of gravity, namely the number +9.8 m/s/s. I used m/s/s instead of m/s² to stress that the acceleration is rate of change of velocity with respect to time.

Warmacblu

Thanks for the explanation.