# Motion in a Plane-Finding velocity and displacement

## Homework Statement

A helium-filled balloon rises straight up, at a rate of 0.7m/s, for a distance of 20m before it pops. A steady breeze of 1.5m/s due east blows against the balloon.

a) What is the velocity of the balloon relative to the ground?
b)How long does the balloon rise before popping?
c) How far from its starting point has the balloon flown(its final displacement)?

→ →
V=Δd/Δt

→ →
Δd=d2-d1

→ →
a= Δv/Δt

Δv=v2-v1

a2+b2=c2

sin=opp/hyp

## The Attempt at a Solution

I am having the most trouble with c but I am also not sure if i did a and b right.

a) We know the balloon is travelling 0.7m/s up and 1.5m/s east. I drew a diagram, and then used pythagorean theorem to solve for the velocity relative to the ground.

(0.7m/s)2+(1.5m/s)2=c2
√2.74=c
c=1.7m/s

b) We know the velocity and the displacement, therefore we can use the velocity formula, rearrange it, and solve for time.

v=Δd/Δt
Δt=Δd/v
Δt= 20m/0.7m/s
t= 28.6s

c) I think i just have to find how far the balloon blows east and then use pythagorean theorem to solve for total displacement?

Thanks so much!

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It seems right to me

Just remember to tell that for question a) velocity is a vector... you found correctly the magnitude, but you should also see the direction... what you could do is compute the angle with respect to the ground (you already have the components so...)

As for c), same as before (you have total time the balloon travels so you can find the horizontal displacement). Also here remember the angle (displacement is also a vector).

ok so the balloon is moving east and up relative to the ground so would i just put [E] for my direction or how would i show the fact that it is also rising?

Final displacement is a vector starting in the take-off point of the balloon and getting to the final position, therefore it is directed up and east... so it will be a vector with some length (found as you told correctly with Pitagora) and directed with an angle with respect to the horizontal. (by the way this angle should be the same as for velocity)

This holds clearly unless they just ask the horizontal displacement... it depends on the interpretation of how far...

To be sure you can compute both

but i cant put [up(degrees)E] so how would i express that?

for [c], the balloon travels 28.6 sec at a velocity 1.7m/sec..........
you have everything you need

if you have any doubts, draw a sketch with time,distance,etc, labelled....

One way is to use polar coordinates $r-\theta$... otherwise draw it

i have drawn it and using the total distance (hyp) and the upward distance (adj) i found the angle with cosine. i found 65.7 degrees as my angle

would my answer just be 48.6m[N66degreesE] ?

Good

Yes, make sure your notation for directions is understood but it is ok

wait...i found the angle for the velocity at the beginning and i got 65 degrees so i should probably just use the 65 degrees seeing as it was calculated with the given numbers and not my calculated rounded ones

48.6m[N65*E] ?

Good, really good work...
(indeed I got 25° computing the other angle using the results not rounded, so 65 is better)

Ok Thanks so much :)

You're welcome