Time Before a Water Balloon Returns

A person throws a water balloon straight up from a height of 2 m with a velocity of 22 m/s. If right under the path of the balloon there is a 1.6 m kid,
How long does the kid have to move to the side in order for him not to be hilted by the balloon?
 a) -0.02 s b) 2.5 s c) -1.5 s N d) 4.5 s e) None of the above.
Solution:

There is a straight form of solving this problem by using the formula  , (F_F 4), which it can be solved for the time when the initial velocity and final position of the balloon are known.

In order for solving for the time in the previous equation, it is necessary to solve the following quadratic equation,

from where the time is

As all quadratic equations, two mathematical solutions have been obtained each of them corresponding to the possible different signs. By understanding the actual physical situation, the actual solution to the problem can be recognized.

The numerical solution is obtained from

where from the two mathematical solutions, the physical solution can be obtained. In this case, the negative solution corresponds to the balloon hitting the kid before it is released by the man. Thus, the only possible solution to the actual physical problem is 4.5 s.

 by Luis F. Sáez, Ph. D. Comments and Suggestions: LSaez@dallaswinwin.com