Understanding Trigonometry and Physics in Balloon Flight

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Homework Help Overview

The discussion revolves around the application of trigonometry and physics concepts to analyze the motion of a balloon rising vertically while being affected by a horizontal breeze. The problem involves calculating the balloon's velocity relative to the ground and understanding the components of its motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the identification of the hypotenuse and the sides of the triangle formed by the balloon's motion. Questions arise regarding the definitions of opposite and adjacent sides in relation to angles. There is also exploration of using Pythagoras' theorem to find the resultant velocity and displacement.

Discussion Status

Participants are actively engaging with the problem, sharing insights about the components of velocity and the implications of time on horizontal distance. Some guidance has been provided regarding the use of vector components and the relationship between time and distance, though there remains a lack of explicit consensus on certain definitions and interpretations.

Contextual Notes

There is mention of potential confusion regarding the orientation of axes in a three-dimensional context, particularly how vertical motion relates to cardinal directions. Participants note discrepancies between textbook definitions and their own interpretations of the problem setup.

physicsgal
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i basically know what to do.. but i have problems drawing the triangle and knowing what sides are which.

"a balloon rises upward, at a rate for 0.7 m/s, to a height of 20m before it pops. a breeze of 1.5m/s due east blows against the balloon."

so the hypotenuse (sp?) is the unknown side?
and how does one know which angle is the opposite angle?

~Amy
 
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hypotenuse would be the unknown side, yes. But all you need is Pytagoras' theorem. you don't need to know angles.
 
thanks quasar

so the hypotenuse side is 20.056m.

the question goes on to say "what is the velocity of the balloon relative to the ground?"

so i need to know which one of the opposite and adjacent sides.

~Amy
 
Nah, again, you just need to identify the components of the velocity vectors [itex]\vec{v}[/itex] and compute its norm, which is again, applying pythagoras' theorem to find the hypotenuse of the triangle whose other 2 sides are the components of the vector.
 
two other questions for this are:

"how long does the balloon rise before popping?"

and "how far from its starting point has the balloon flown (its final displacement)?

i probably need to know what's the opposite/adjacent side.

~Amy
 
(the hypotenuse side is not 20.056m)

The balloon has a vertical speed of 0.7 m/s and a horizontal speed of 1.5 m/s. So do you agree that its motion will be diagonal? At the end, it will have moved vertically 20 m and some distance 'X' horizontally. The total diagonal displacement D will be given by Pythagoras' theorem: D=[itex]\sqrt{20^2+X^2}[/itex]. It's your job, using the equation of kinematic, to find what X is.

The first step towards this is to find what time it takes the balloon to reach its height of 20 m at 0.7 m/s. The knowledge of this time will then allow you to find what horizontal distance X the balloon traveled btw the time when the balloon was on the ground and the time when it had reached its height of 20 m.
 
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1) since the two components of velocity are already given
I would just repeat them . . . v = (1.5, 0, 0.7)m/s as (E,N,up).
Once you know the components, I wouldn't use Pythagoras to hide them!

1b) If you are asked specifically for magnitude and direction angle,
. then YOU get to decide which side is opposite and which is adjascent
Draw the vector components as arrows tail-to-tip,
and draw the diagonal from first tail to last tip. The triangle has
one skinny angle and one wide angle. If you mark an angle with an arc,
the arc will touch one of the sides as well as the diagonal.
THAT side is adjascent to THAT angle.
A side is OPPOSITE the angle whose arc does NOT touch it.

2) once you find the time to rise 20 m at 0.7 m/s , use that time
to find the horizontal distance . . . you might notice that the triangle has
a straight line, and you COULD use a proportion . . . avoid the temptation ! the procedure of connecting two dimensions via time is widely applicable.
 
thanks for the tips!

here's what i have:
a) 1.66 m/s (N 25 degrees E)

b) H = 20m/sin 25 degrees
= 47.32 m

47.32 m/1.66 m/s = 28.5s

c) (28.5s)(1.66m/s)
= 47.3 m.

i hope everythings accurate..

also, in my text i was just reading that the unknown angle is the one beside the starting point.

~Amy
 
Technically, the answer to a) is 25 degrees from the horizontal. You gave a bearing, which lies in a different plane.
 
  • #10
Technically, the answer to a) is 25 degrees from the horizontal. You gave a bearing, which lies in a different plane.

im not sure what you mean :blushing:

but everything else looks ok?

~Amy
 
  • #11
physicsgal said:
im not sure what you mean :blushing:

but everything else looks ok?

~Amy
North isn't in the vertical plane, in this case it could be in the z plane. Imagine drawing 3D axes. The y-axis represents up and down and the x-axis represents the horizontal. The z axis could represent north. Your balloon is rising upwards not going north.

Does that make sense?
 
  • #12
North isn't in the vertical plane, in this case it could be in the z plane. Imagine drawing 3D axes. The y-axis represents up and down and the x-axis represents the horizontal. The z axis could represent north. Your balloon is rising upwards not going north.

thanks. i see what you're saying. going up/down doesn't mean north/south. but in my lesson back it kind of suggest up is north, and down is south. i dunno. i guess they are just doing that for simplicity.

~Amy
 

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