Vertical motion, gravity question?

Question

A hot air balloon is 80 m above the ground and travelling vertically downards at 8.0 m/s when one of the passengers accidently drops a coin over the side. How long after the coin reaches the ground does the balloon touch down?


Ok so i assume ur supposed to find the landing times of both. So for the hot air balloon, since it's travelling at 8m/s and there's 80 metres to go, it'll take 10 seconds to reach ground. Is that right?

Then for the coin i can't figure out how to get the time. Which equation of motion do you use?
u = 8m/s
x = 80 or is it -80?
a = 9.8m/s^2

do you use x = ut + .5at^2? That's the only equation with all the values that I have but then you'd have to work out 't' as a quadratic equation.

I dunno maybe i'm getting the theory wrong and just not understanding this topic.

Any ideas on how to do it?


Thanks!!

Answer 1

I'm assuming that the balloon is descending at a constant speed.
t(balloon)=t1= y/v=80/8=10s
t(coin)= t2
then
y= vt2 + 0.5gt2^2
0.5g(t2)^2 + vt2 -y=0

t2=[- v + sqrt( v^2 + 4y(0.5g)]/(g)
t2=[-8+ sqrt( 8^2 + 2x80(9.8)]/(9.8)
t2=3.3s

sqrt(2y/g)=sqrt(2x10/9.81)=1.4 s

t=t1-t2 is what we need to find

t= 10 -3.3 =6.7s

Answer 2

You are right about the balloon. But the coin will accelerate towards the ground. I can't remember the acceration rate in metric - 32 ft per second per second in Imperial! It's starting velocity will be zero and then accelerate to the ground.

Answer 3

All I can say is that the coin accellerates down at 32 feet per second squared ( gravity).i.e. 32 feet for the first second, 64 feet in the second second, a 128 feet in the third etc.
You do the conversions. I was brought up on Imperial Measure

Answer 4

s=1/2at^2+ut
s=80 m
a=9.8
u=8ms^-1
t=??
80=1/2(9.8)(t)^2+(8)(t)

t=3.305920383 s.

now Ballon takes 10 seconds to reach the ground .

10-3.305920383 = 6.694079617 s.

Answer 5

you realy beleve that someone will read that?