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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

A balloon takes off the ground and rises at a rate of 20 ft per second. after 10 seconds a ball is dropped out of the balloon, how many seconds after the ball is dropped will it hit the ground?

For the sake of the problem, assume the balloons initial height is zero. The height of the ball equals the height of the balloon untill it is let go, and the acceleration of gravity is exactly -32 ft/sec

Please do not display how you worked out the problem as not to spoil it for others. Just provide your answer. Remember, the answer is not how many seconds after the balloon is released, but how many seconds from when the ball is let go, untill it hits the ground.

Good luck!

*Last edited by mikau (2006-01-12 07:39:10)*

A logarithm is just a misspelled algorithm.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I think it's this:

Why did the vector cross the road?

It wanted to be normal.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Correct! Nice job mathsy! But you can display the answer so long as you don't show how you worked it.

Dang. You got it first try. Lets me rewrite the problem to make it less obvious. (edit) okie doke. Lets see if anyone else can get it. (note, the answer has changed, mathsy's answer was correct for the original problem)

*Last edited by mikau (2006-01-12 07:40:28)*

A logarithm is just a misspelled algorithm.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

4.215 seconds

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Correct! Boy you guys are smart.

My dad told me about this problem. He got it in highschool. Everyone got it wrong because everyone assumed the balls initial velocity was 0 at the instant it is let go, but actually, at that instant, the balls velocity equals the velocity of the balloon! Usually people miss this but neither of you did. Well done!

I love calculus....

A logarithm is just a misspelled algorithm.

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