How Far Should a Helicopter Drop a Food Packet for Flood Victims?

[gracy]

A helicopter on a flood relief mission flying horizontally with a speed vo at an altitude H, has to drop a food packet for victims standing on the ground. At what 'distance from the victims should the food packet be dropped'?the victims are standing in the vertical plane of the helicopter.
In my textbook diagram for this problem is given
(a)

but what i am confused about is why victims are standing there, can't they stand just beneath the helicopter like this (b)

after all ,victims are standing in the vertical plane of the helicopter in this picture also .So 'distance from the victims should the food packet be dropped 'is equal to height H .That's what i thought and chose the option H as answer for this MCQ.But in my textbook 'distance from the victims should the food packet be dropped' i.e D is taken as in Picture (a) and hence solved this problem in totally different way.So what is wrong in picture b ,victims are still in the vertical plane of the helicopter as mentioned in the question.please do reply.
Thanks.

 

[Doc Al]

When dropped from a moving helicopter (looks like a plane to me), do the packets fall straight down?

 

[gracy]

When dropped from a moving helicopter (looks like a plane to me), do the packets fall straight down?

assume it as helicopter.

 

[gracy]

do the packets fall straight down?

I didn't understand what you are trying to ask.please clear your question

 

[Doc Al]

I didn't understand what you are trying to ask.please clear your question

I'll ask in a different way: When the packets first leave the helicopter, what velocity do they have?

 

[gracy]

I'll ask in a different way: When the packets first leave the helicopter, what velocity do they have?

it is not given.i am just concerned and confused about position of victims.please review my full question.

 

[DocZaius]

I think you are confused by the fact that a helicopter is mentioned. It seems like since the problem statement mentions a helicopter and you don't see a helicopter in the figure, you are ignoring other parts of the figure (such as the fact that the vehicle in question is not hovering over the survivors). I think it'd be reasonable for you to assume that the helicopter/plane is not hovering over the survivors. Then this becomes a non-trivial (in that you have to use at least one equation) projectile motion problem.

I recommend you use figure a. Assume the vehicle is some lateral distance away from the survivors when it drops the package.

 

[Doc Al]

it is not given.

Actually, it is given.

i am just concerned and confused about position of victims.please review my full question.

I understand your question. Please answer my questions, as they are meant to clear up yours.

(And read DocZaius's post above.)

 

[gracy]

I think you are confused by the fact that a helicopter is mentioned. It seems like since the problem statement mentions a helicopter and you don't see a helicopter in the figure, you are ignoring other parts of the figure (such as the fact that the vehicle in question is not hovering over the survivors). I think it'd be reasonable for you to assume that the helicopter/plane is not hovering over the survivors. Then this becomes a non-trivial (in that you have to use at least one equation) projectile motion problem.

I recommend you use figure a. Assume the vehicle is some lateral distance away from the survivors when it drops the package.

i think there is some misunderstanding.picture (a)is given in my textbook but not in question but as a solution.

 

[gracy]

Actually, it is given.

how?initial velocity of packet is same as that of helicopter i.e vo,right?

 

[lep11]

how?initial velocity of packet is same as that of helicopter i.e vo,right?

yes

 

[BiGyElLoWhAt]

how?initial velocity of packet is same as that of helicopter i.e vo,right?

which is a velocity, represented as an arbitrary unknown rather than a specific number.

 

[DocZaius]

i think there is some misunderstanding.picture (a)is given in my textbook but not in question but as a solution

Ah ok. In that case I understand your confusion more. However the problem statement does say the helicopter is flying horizontally. This means that the package will be moving both horizontally and vertically upon being dropped. So the helicopter cannot be directly above the survivors at the moment it drops the package. In your (second) diagram, the package would miss the survivors since it would move horizontally at the speed of the vehicle. The vehicle needs to be located laterally "before" the survivors. Find that distance.

 

[Doc Al]

i think there is some misunderstanding.picture (a)is given in my textbook but not in question but as a solution.

The picture fits the question (except they show a plane, not a helicopter).

how?initial velocity of packet is same as that of helicopter i.e vo,right?

Right. So what sort of trajectory would you expect it have as it fell? Would you expect it to fall straight down?

 

[gracy]

This means that the package will be moving both horizontally and vertically upon being dropped.

why?can you please make me understand this point.

 

[Doc Al]

why?can you please make me understand this point.

Drop a ball. How does it fall to the ground?

Throw a ball horizontally. How does it fall to the ground?

 

[gracy]

Drop a ball. How does it fall to the ground?

Throw a ball horizontally. How does it fall to the ground?

in first case vertically and in second case it's path would be projectile.

 

[Doc Al]

in first case vertically and in second case it's path would be projectile.

Right. The second case would take a parabolic trajectory. (Both cases are projectile motion.)

The second case is exactly like the one in your problem. Since the packet has an initial horizontal velocity, it is just as if it were thrown at that speed.

 

[gracy]

how first case is an example of projectile motion?

 

[BiGyElLoWhAt]

better example than throwning a ball. Drop the ball. Have your friend get in a car and drive past your house and then drop the ball. The second case is identical to the situation, except it's a helicoptor (plane?), not a car.

 

[BiGyElLoWhAt]

how first case is an example of projectile motion?

The projectile is being "projected" directly downwards. It doesn't have to be shot sideways out of a cannon to be a projectile.

 

[gracy]

The projectile is being "projected" directly downwards. It doesn't have to be shot sideways out of a cannon to be a projectile.

actually i have tried this at home.ball fall vertically downward not followed parabolic path as in projectile motion.

 

[gracy]

Drop a ball.

actually i have tried this at home.ball fall vertically downward not followed parabolic path as in projectile motion.

 

[cjl]

actually i have tried this at home.ball fall vertically downward not followed parabolic path as in projectile motion.

I'm guessing you weren't moving when you dropped the ball then. Now drop the ball while walking. Does it still fall vertically?

EDIT: Also, as BiGyElLoWhAt points out below, it's still following a path that is described as "projectile motion". A vertical free-fall is a special case of projectile motion with no sideways velocity.

 

[BiGyElLoWhAt]

It doesn't have to be parabolic to be projectile. It's a projectile with zero x component in the velocity.

 

[BiGyElLoWhAt]

I'm guessing you weren't moving when you dropped the ball then. Now drop the ball while walking. Does it still fall vertically?

You won't notice it walking. Get on a bike or a skateboard or have a friend drive a car and just watch the ball. Maybe pull out your phone and record it, but make sure to keep the phone stationary when you do. Watch the ball leave your friends hand and shoot across the screen out of view of the camera.

 

[gracy]

. A vertical free-fall is a special case of projectile motion with no sideways velocity.

you mean straight vertical path is also a projectile?i have never been told this in class.

 

[cjl]

Yes it is - it follows the exact same equations of motion. Try it - if you use whatever equations you were given in class, and plug in zero for the initial horizontal velocity, what do you get?

 

[gracy]

- it follows the exact same equations of motion.

equations of motions?suvat equations,right?we will not take projectile specific equations such as for maximum height,range,time of flight in this type of vertical free fall projectile motion?

 

[cjl]

All of those equations work just as well for vertical projectile motion as they do for angled projectile motion. As I said - try using them with some arbitrary initial conditions (with zero initial horizontal velocity) and see what they tell you.

 

[BiGyElLoWhAt]

equations of motions?suvat equations,right?we will not take projectile specific equations such as for maximum height,range,time of flight in this type of vertical free fall projectile motion?

Do it all if you want. It all works. In a vertical free fall (starting from rest) y_0=0 and x_0=0. Plug it in. I think it would be good to work out.

Edit: ^ That response time, though.:D

 

[cjl]

Edit: ^ That response time, though.:D

:D

 

[gracy]

starting from rest

with zero initial horizontal velocity)

with zero initial horizontal velocity or starting from rest i.e both vertical as well as horizontal velocity should be taken as zero for applying projectile motion equations to vertical free fall projectile motion?

 

[cjl]

You can do either. You can have the object start from rest at some initial height h₀, or you can have it start with some initial height h₀ and some initial vertical velocity V_y,0. Either way, you should get purely vertical motion.

 

[gracy]

You can have the object start from rest

but if i will take initial velocity as zero all -maximum height,range,time of flight would come out as zero because all of them have initial velocity in their formula.

 

[BiGyElLoWhAt]

but if i will take initial velocity as zero all -maximum height,range,time of flight would come out as zero because all of them have initial velocity in their formula.

No it doesn't. Write the equations of motion please.

 

[cjl]

Not if you start at some nonzero initial height h₀.

 

[gracy]

No it doesn't. Write the equations of motion please.

equation for Range =u square sin2theta/g
time of flight=2u sin theta/g
maximum height=u square sin square theta/2g
here u=initial velocity
what i am missing?
is u not equal to initial velocity.then u is what ?

 

[cjl]

Are those the only form of the equations you were given (or have in your book)? Those implicitly assume a starting height of zero, which is incorrect both for the quick example that BiGyElLoWhAt and I are trying to get you to work, and for the problem stated in the original post.

 

[gracy]

Those implicitly assume a starting height of zero, which is incorrect

is it incorrect or correct but not the only case of the projectile?

 

[cjl]

Your equations are correct for a special case (namely, where the starting height is zero), which does not apply to the problem as stated in the original post (or the problem we want you to work out).

 

[gracy]

Are those the only form of the equations you were given (or have in your book)? Those implicitly assume a starting height of zero, which is incorrect both for the quick example that BiGyElLoWhAt and I are trying to get you to work, and for the problem stated in the original post.

oh i got your point .give me some time i will work it out by myself.

 

[BiGyElLoWhAt]

I have no idea what you're writing. The position for an object at any point in time (with constant acceleration) is given by:
$x= x_0 + v_{(0\ of\ x)} t + \frac{1}{2}a_xt^2$
y is the same, but with y's.

 

[gracy]

I have no idea what you're writing. The position for an object at any point in time (with constant acceleration) is given by:
$x= x_0 + v_{(0\ of\ x)} t + \frac{1}{2}a_xt^2$
y is the same, but with y's.

i am sorry but i didn't understand your post.

 

[cjl]

i am sorry but i didn't understand your post.

Do you have any other equations than the ones you mentioned above? As I said, the ones you already posted won't work for what we want to do right now.

 

[gracy]

Do you have any other equations than the ones you mentioned above? As I said, the ones you already posted won't work for what we want to do right now.

no,i don't have.i have given and taught these only.i want to learn and explore more.. i tried.as far as i know
the range is defined as the distance between the launch point and the point where the projectile hits the ground.so in this vertical free fall projectile it (range) should be simply height from where the ball is dropped.right?

 

[cjl]

Range is defined as the horizontal distance from launch point to impact point, so for a vertical free fall, range is zero. Was the problem in the original post given to you in class?

 

[gracy]

Was the problem in the original post given to you in class?

original post?that helicopter problem?

 

[cjl]

Yes, that one.

 

[gracy]

Yes, that one.

yes,it is part of my physics practice worksheet.but it is not about vertical free fall projectile.