Asked by Nick
A ballon rises at 14m/s, a boy in the ballon throws a ball straight up at 24m/s. How much later does he catch the ball?
Answers
Answered by
bobpursley
In the balloon?
What does the baloon rising have to do with it?
Now the balloon is still rising...
the balloon keeps rising...
distance the balloon travels upward from the start to the finish.
d=14*tup + 14*tdown.
Where does the ball land? at d.
14(timetotal)=38*t-1/2 g t^2
or 14=38-1/2 gt or
t=2(38-14)/g = 2(24)/g
Now consider that the balloon movement has nothing to do with it.
d=0=24*t-1/2 g t^2
or t= 2*24/g
The logical conclusion is that if the boy in the balloon didn't consider that he was moving (it was at night), the problem is the same moving as no moving.
So, a final question to consider: what if the balloon were accelerating upward?
You can work that out.
What does the baloon rising have to do with it?
Now the balloon is still rising...
the balloon keeps rising...
distance the balloon travels upward from the start to the finish.
d=14*tup + 14*tdown.
Where does the ball land? at d.
14(timetotal)=38*t-1/2 g t^2
or 14=38-1/2 gt or
t=2(38-14)/g = 2(24)/g
Now consider that the balloon movement has nothing to do with it.
d=0=24*t-1/2 g t^2
or t= 2*24/g
The logical conclusion is that if the boy in the balloon didn't consider that he was moving (it was at night), the problem is the same moving as no moving.
So, a final question to consider: what if the balloon were accelerating upward?
You can work that out.
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