Kinematics: Catch Ball Dropped from Balloon After 24.7s

[negation]

Homework Statement

A balloon is rising at 10m/s when its passenger throws a ball straight up at 12m/s relative to the balloon . How much later does the passenger catches the ball?

Homework Equations

The Attempt at a Solution

vf^2 - vi^2 = 2g(yf-yi)

Ball:
0ms^-1 - (22ms^-1)^2 = 2(-9.8ms^-2)(yball)
yball = 24.7m

Balloon:
0ms^-1 - (10ms^-1)^2 = 2(-9.8ms^-2)(yballoon)
yballoon = 24.7m

 

[voko]

Assume that the velocity of the balloon is constant.

 

[negation]

Assume that the velocity of the balloon is constant.

You're right since the ball is thrown up relative to the balloon.

 

[voko]

The balloon is not a projectile, there are other forces acting on it. That does not mean, in general, that the acceleration is zero, but since you are not given any, the only sane assumption is that the acceleration is zero.

 

[negation]

yf - yi = vit + 0.5gt^2

Ball:

yball = (12ms^-1 + 10ms^-1)t + 0.5(-9.8ms^-2)t^2

Balloon:

yballoon = 10ms^-1 t + 0.5(oms^-2)t^2
y balloon = 10ms^-1

yballoon = yball

10ms^-1 t + 0.5(oms^-2)t^2 = 10ms^-1
t = 2.4s

 

[voko]

There is a simpler approach. Since the balloon is not accelerated, everything can be computed with regard to the balloon as if it was stationary. So that basically means the projectile goes up at 12 m/s. At some time t it will reach a maximum elevation (from the balloon). This time can be easily computed from the condition that the (relative) velocity becomes zero. Going down takes the same time.

 

[negation]

There is a simpler approach. Since the balloon is not accelerated, everything can be computed with regard to the balloon as if it was stationary. So that basically means the projectile goes up at 12 m/s. At some time t it will reach a maximum elevation (from the balloon). This time can be easily computed from the condition that the (relative) velocity becomes zero. Going down takes the same time.

So is 2.4s the right answer?
It appears more logical if I multiply 2.4s by 2.

 

[haruspex]

So is 2.4s the right answer?
It appears more logical if I multiply 2.4s by 2.

You correctly calculated 2.4s as the time when ball and balloon would meet again. If you had calculated the time to when the relative velocity were zero, then you would have got 1.2s and would need to double for the second half.
In fact, a slightly simpler calculation than the one you did is to say that they will meet again when the relative velocity is equal and opposite to its initial value, giving gt = 24m/s.