Finding Time, Distance, and Velocity

[Heat]

Homework Statement

A military helicopter on a training mission is flying horizontally at a speed of 60.0 m/s and accidentally drops a bomb (fortunately not armed) at an elevation of 300m. You can ignore air resistance.

How much time is required for the bomb to reach the earth?
How far does it travel horizontally while falling?
Find the horizontal and vertical component of its velocity just before it strikes the earth.

Homework Equations

Velocity Final = Velocity Initial + Acceleration(Time)
X Final - X Initial = Time/2 (Velocity Initial + Velocity Final)
X Final - X Initial = Velocity Initial (Time) + .5 (Acceleration)(Time)^2
Velocity Final^2 = Velocity Initial ^2 + 2 ( Acceleration)(X Final - X Initial)

The Attempt at a Solution

I want to really understand this, so I will be breaking up the work into parts. As usually the result of the previous question help solve the one after it.

For the first question:
How much time is required for the bomb to reach the earth?

The knowns: Final Position @ 300 and Initial @ 0. Acceleration @ -9.8. Velocity Initial @ 60.
The unknown: Time

Based on the values provided, the following equation seems suited to the problem.

X Final - X Initial = Velocity Initial (Time) + .5 (Acceleration)(Time)^2

0 - 300 = 60(t) + .5 (-9.8)(t^2)
-4.9t^2 +60t +300 = 0

-60 +- 97.365 / -9.8

t = 16.06
or
t = -3.81

This incorrect. How would I have to approach this problem.

 

[Kurdt]

You have to think about this problem in two dimensions. To find the time you are along the correct lines. You know it will experience a constant acceleration until it hits the ground 300 m below. But the vertical component of velocity for the bomb when it is released is 0 m/s not 60 m/s.

Also remember your units.

 

[Heat]

So the knows would have to be:

The knowns: Final Position @ 300 and Initial @ 0. Acceleration @ -9.8. Velocity Initial @ 0

0 - 300 = 0(t) + .5 (-9.8)(t^2)
-4.9t^2 = -300
t^2 = 61.22
t= 7.82

also, since it's 300 below , would it not be -300?

 

[Kurdt]

Yes that looks ok. Now how about part two.

 

[Kurdt]

also, since it's 300 below , would it not be -300?

Sorry missed this bit.

It all depends how you are defining things. All that matters is that you are consistent. You could for instance say the ground is at 0m and the helicopter is at 300m or the other way round. Or the helicopter is at 0m and the ground is -300 m.

 

[Heat]

yes, but here I am placing the "origin" at the helicopter, so the ground is -300 downwards. Hence why I am confused why the equation does not look like this instead:

0 "+" 300 = 0(t) + .5 (-9.8)(t^2)

 

[Kurdt]

yes, but here I am placing the "origin" at the helicopter, so the ground is -300 downwards. Hence why I am confused why the equation does not look like this instead:

0 "+" 300 = 0(t) + .5 (-9.8)(t^2)

Well in that case you'd have to define the acceleration downwards as a positive quantity otherwise the equation won't make much sense.

 

[Heat]

How far does it travel horizontally while falling?

The knowns: Final Position @ 300 and Initial @ 0. Acceleration @ -9.8. Velocity Initial @ 0. Time final = 7.82.

Horizontally means the x axis.

If the object falls, would it not just fall down straight. (maybe the little push from the copter)

Either way:

X Final - X Initial = Time/2 (Velocity Initial + Velocity Final)
X Final - 0 = 7.82/2 (60+ 0)
?

 

[Kurdt]

The initial velocity in the horizontal direction is 60 m/s and so is the final velocity. The reason this doesn't change is because the question asks us to ignore air resistance.

 

[Heat]

Well just a bit ago, I realized that horizontal velocity is 60, but why is the final velocity 60 still. I would think that once it bombards into the ground, velocity would be 0.

 

[Kurdt]

Well just a bit ago, I realized that horizontal velocity is 60, but why is the final velocity 60 still. I would think that once it bombards into the ground, velocity would be 0.

Yes once it hits the ground it will be zero, but the equations you are using will assume that it undergoes some uniform decelleration if you assume the final velocity is zero which it doesn't. So we say it hits the ground in 7.82s in which time its traveling at a constant horrizontal speed of 60m/s.

 

[Kurdt]

yes, but here I am placing the "origin" at the helicopter, so the ground is -300 downwards. Hence why I am confused why the equation does not look like this instead:

0 "+" 300 = 0(t) + .5 (-9.8)(t^2)

Just going back to this for a second. That would be incorrect, since the equations you are using state x_{final} - x_{initial} = ... and thus you'd get -300 + 0 = -300.

 

[Heat]

so it would be more like this:

X Final - 0 = 7.82/2 (60+ 60)
X Final - 0 = 7.82/2 (120)
469.2 = x final

 

[Kurdt]

so it would be more like this:

X Final - 0 = 7.82/2 (60+ 60)
X Final - 0 = 7.82/2 (120)
469.2 = x final

That looks correct.

 

[Heat]

ok so we know that the horizontal component prior to reaching ground is 60m/s. and that the vertical when prior to reaching ground is 0 right..

nvm the vertical

i will need to solve using difference in distance over time.

 

[Kurdt]

Just use the first equation you have in the first post to find the vertical component before impact. You know the initial velocity is 0 m/s.

 

[Heat]

the first equation I have in the first post was : Velocity Final = Velocity Initial + Acceleration(Time)

how does this determine max height?

anyways here it goes

vf = 0 + -9.8(3.41)

-33.418...

hehehe...I noticed that I mixed it with another problem.

 

[Kurdt]

I thought the last question was concerned with what the horizontal and vertical components of velocity were. I also thought we had determined the time to be 7.82 s.

 

[Heat]

yes time is 7.82 for the bomb to reach earth, and we are trying to find the horizontal and vertical components of velocity of bomb just before it hits earth. You helped me determine that the horizontal is 60 m/s.

We are trying to figure out the vertical.

and you also pointed out to use the first equation...

I just did, and I get the result of -76.63

v=-9.8 (7.82)

redoing a^2+b^2 =c^2

a^2 + 60^2 = 76.63^2
a= 47.68

it gives me 47.68

which is wrong. :( only got one more try. :(

 

[Kurdt]

I don'y understand what you've done. The question asks for what the components are. You seem to be trying to work out the final velocity before impact. Are you sure the question is as you have stated it?

 

[Heat]

The question as it's stated is the following:

A military helicopter on a training mission is flying horizontally at a speed of 60.0 m/s and accidentally drops a bomb (fortunately not armed) at an elevation of 300 m. You can ignore air resistance.

Find the horizontal and vertical component of its velocity just before it strikes the earth.

 

[learningphysics]

The question as it's stated is the following:

A military helicopter on a training mission is flying horizontally at a speed of 60.0 m/s and accidentally drops a bomb (fortunately not armed) at an elevation of 300 m. You can ignore air resistance.

Find the horizontal and vertical component of its velocity just before it strikes the earth.

You've already found the answers. Horizontal compnent = 60m/s Vertical component = -76.63 m/s

 

[Heat]

Ok, I thought I would be able to do this using the formulas from prior, but I don't understand it:

Draw x-t graph for the bomb’s motion.

http://img221.imageshack.us/img221/8004/untitlednw6.jpg

also tried initially to set at time =0 the x at 300 and go down, but that did not result either. :(

fot this I used x final -300 = 60t +.5 ( -9.8)t^2

I also need to draw y-t graph for the bomb’s motion. Which I would think is the same as the x graph.

as the bombs moves along the horizontal, the vertical will down. it should be one of the "C" (just cut off the top part of the c) :)

I already know that the graph of the x velocity versus time is just a constant line going through the velocity mentioned.

The same would apply to vertical velocity but at -9.8s.

 

[Heat]

also in addition to the question in my last post:

What if the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground?

I would say it still remains in flight as that is where it was at the beginning.

 

[learningphysics]

The important thing about these questions is to be able separate the horizontal aspect from the vertical aspect... deal with them separately...

The horizontal velocity of the helicopter is constant. It starts at 60m/s... So what is x(t)?

What you plotted in your graph is actually y(t).

 

[Kurdt]

I'm not sure exactly where you are having difficulty with these concepts. If you could say how you don't understand or what bits then perhaps we can help.

 

[Heat]

I understood, and graphed them successfully. :)

 

[Bottomsouth]

You've already found the answers. Horizontal compnent = 60m/s Vertical component = -76.63 m/s

I understand how you get the horizontal component but how did you arrive to the vertical one.

-300 m cleared in 7.82 s is -38.3 m/s. Now -38.3 x 2 = -76.6.

But what equation did you use?