Will Supplies Dropped from a Plane Reach an Isolated Flooded Community?

[-Dragoon-]

Homework Statement

A pilot is attempting to deliver emergency good and first-aid supplies to an isolated northern community that has suffered severe flooding. The plane has a horizontal velocity of 1.40 X 10^2 km/h, as it flies at an altitude of 1.80 X 10^2 metres. The community is situated on a dry patch of land that is about 72m X 72m.
a) If the supplies are released just as the plane flies directly overhead, will they touch down on land or in water? Justify your response with calculations that show exactly where the package will land.
b) When should the supplies be released so that they touch down very close to the centre of the dry patch of land? Answer in terms of distance from the target, not time.

Homework Equations

Vertical distance = 1/2(acceleration)(delta t^2)
Horizontal distance = (horizontal velocity)(delta t)
c^2 = a^2 + b^2

The Attempt at a Solution

First I drew this diagram to help me along the way: http://img683.imageshack.us/i/projectilemotion.png/

a) I know that I have the vertical component, but no horizontal component. In order to find the horizontal component, I would need to find the change in time time. I take into account that there is no initial velocity, so I can use the vertical distance = 1/2(acceleration)(delta t^2) rather than Vertical distance =(initial velocity)(delta t)+1/2(acceleration)(delta t^2) equation. Since I'm given vertical distance = 180m, I can isolate and solve for time which is approximately 6.1 seconds. Now I can find the horizontal distance that is traveled:

Horizontal distance = (horizontal velocity)(change in time) = (38.889 m/s after conversion)(6.1) = 235.6 m forward. Now that I have both the horizontal distance it travels and the vertical distance it falls, I can find the total distance the aid will travel before it falls. I use pythagorean theorem to find this, and yield a value of approximately 296.5 m as the resultant distance. Did I do this correctly? This would mean that it will not reach the dry patch of land by a long shot, does it not?

b) This is why I believe I did part a correctly, because the only way the aid will reach the people is if it is released while the plane is at least more than 163 m of water away, due to the horizontal distance that would be traveled by the aid which is 235.6 m forward. Either that, or the pilot would need to lower the velocity of the plane and hence the horizontal traveled will be much shorter.

 

[PhanthomJay]

Homework Statement

A pilot is attempting to deliver emergency good and first-aid supplies to an isolated northern community that has suffered severe flooding. The plane has a horizontal velocity of 1.40 X 10^2 km/h, as it flies at an altitude of 1.80 X 10^2 metres. The community is situated on a dry patch of land that is about 72m X 72m.
a) If the supplies are released just as the plane flies directly overhead, will they touch down on land or in water? Justify your response with calculations that show exactly where the package will land.
b) When should the supplies be released so that they touch down very close to the centre of the dry patch of land? Answer in terms of distance from the target, not time.

Homework Equations

Vertical distance = 1/2(acceleration)(delta t^2)
Horizontal distance = (horizontal velocity)(delta t)
c^2 = a^2 + b^2

The Attempt at a Solution

First I drew this diagram to help me along the way: http://img683.imageshack.us/i/projectilemotion.png/

a) I know that I have the vertical component, but no horizontal component. In order to find the horizontal component, I would need to find the change in time time.

change the word 'component' to 'displacement' , or 'component of displacement'

to I take into account that there is no initial velocity, so I can use the vertical distance = 1/2(acceleration)(delta t^2) rather than Vertical distance =(initial velocity)(delta t)+1/2(acceleration)(delta t^2) equation. Since I'm given vertical distance = 180m, I can isolate and solve for time which is approximately 6.1 seconds. Now I can find the horizontal distance that is traveled:

Horizontal distance = (horizontal velocity)(change in time) = (38.889 m/s after conversion)(6.1) = 235.6 m forward.

yes, correct

Now that I have both the horizontal distance it travels and the vertical distance it falls, I can find the total distance the aid will travel before it falls. I use pythagorean theorem to find this, and yield a value of approximately 296.5 m as the resultant distance. Did I do this correctly? This would mean that it will not reach the dry patch of land by a long shot, does it not?

Yes, but you don't have to calculate the resultant displacement, just the horizontal displacemnt is all you need.

b) This is why I believe I did part a correctly, because the only way the aid will reach the people is if it is released while the plane is at least more than 163 m of water away, due to the horizontal distance that would be traveled by the aid which is 235.6 m forward. Either that, or the pilot would need to lower the velocity of the plane and hence the horizontal traveled will be much shorter.

The problem asks for the distance from the center of the patch, but you have the right idea.

 

[-Dragoon-]

^ Okay, so all I need is the horizontal displacement and the vertical displacement not being taken into account? But wouldn't that mean the aid does not have a vertical displacement, when it clearly does?

and for part b, I didn't really take into the centre of the patch which about 36m. Wouldn't that mean the pilot has to drop the aid 199.6 m away from the land taking into account that all wee need is the horizontal displacemnt (235.6m - 36m)?

 

[PhanthomJay]

^ Okay, so all I need is the horizontal displacement and the vertical displacement not being taken into account? But wouldn't that mean the aid does not have a vertical displacement, when it clearly does?

well it has both, of course, but it depends on how the pilot measures the distance to the target. If along the diagonal, then yes, you need to know that, but if instrumentation measures horizontal distance from a target, that's all you need.

and for part b, I didn't really take into the centre of the patch which about 36m. Wouldn't that mean the pilot has to drop the aid 199.6 m away from the land taking into account that all wee need is the horizontal displacemnt (235.6m - 36m)?

Well, no, the pilot must release it 235 m horizontally from the center of the patch (the target), or between 200 and 270 m horizontally away from the target in order for the aid to land somewhere on the patch.

 

[-Dragoon-]

well it has both, of course, but it depends on how the pilot measures the distance to the target. If along the diagonal, then yes, you need to know that, but if instrumentation measures horizontal distance from a target, that's all you need.

Erm I still don't get it...
In what situation would they ask for the resultant distance (displacement), and when would I just need the horizontal displacement? That is exactly what I'm struggling to understand. For example, I know that I would only need vertical displacement only when the pilot has no velocity and is just cruising above, but it's not at least what I think that he's flying his plane on the ground which would then mean there is no vertical displacement. Could you just give an example where I would need the resultant, and one where I wouldn't? Thanks in advance for all the help, by the way.

Well, no, the pilot must release it 235 m horizontally from the center of the patch (the target), or between 200 and 270 m horizontally away from the target in order for the aid to land somewhere on the patch.

Okay, I understand this clearly. Thank you.

 

[PhanthomJay]

Erm I still don't get it...
In what situation would they ask for the resultant distance (displacement), and when would I just need the horizontal displacement? That is exactly what I'm struggling to understand. For example, I know that I would only need vertical displacement only when the pilot has no velocity and is just cruising above, but it's not at least what I think that he's flying his plane on the ground which would then mean there is no vertical displacement. Could you just give an example where I would need the resultant, and one where I wouldn't?

The vertical and horizontal displacements are just the components of the resultant displacement. In your problem, from release to impact, the aid is displaced 180 m vertically and 235 m horizontally, or, as you noted, it has a total resultant displacement of 296 m (at an angle of about 37.5 degrees below the horizontal). They are each a different way to say the same thing. During this same time period, the pilot is displaced 235 m horizontally and 0 m vertically, so his displacement is 235 m in the horizontal direction (he ends up directly overhead the aid at the time of impact).

What you don't want to do is confuse distance, which is a scalar quantity withot any direction, with displacement, which is a vector quantity with direction. There is no such thing as a 'resultant' distance. Note that in this problem, the magnitude of the aid's displacement , 296 m, is less than the distance it travels, because the distance would be measured along its parabolic curve, which is greter than the diagonal distance along the displacemnt vector.

 

[-Dragoon-]

The vertical and horizontal displacements are just the components of the resultant displacement. In your problem, from release to impact, the aid is displaced 180 m vertically and 235 m horizontally, or, as you noted, it has a total resultant displacement of 296 m (at an angle of about 37.5 degrees below the horizontal). They are each a different way to say the same thing. During this same time period, the pilot is displaced 235 m horizontally and 0 m vertically, so his displacement is 235 m in the horizontal direction (he ends up directly overhead the aid at the time of impact).

What you don't want to do is confuse distance, which is a scalar quantity withot any direction, with displacement, which is a vector quantity with direction. There is no such thing as a 'resultant' distance. Note that in this problem, the magnitude of the aid's displacement , 296 m, is less than the distance it travels, because the distance would be measured along its parabolic curve, which is greter than the diagonal distance along the displacemnt vector.

Ah, I get it. Thanks for all the help.