#### Answer

The ball reaches a height of 3.08 meters above the top of the tube.

#### Work Step by Step

We can find the acceleration of the ball while it is in the tube;
$\sum F = ma$
$F_{air}- mg = ma$
$a = \frac{F_{air}- mg}{m}$
$a = \frac{2.0~N- (0.050~kg)(9.80~m/s^2)}{0.050~kg}$
$a = 30.2~m/s^2$
We can find the speed as the ball leaves the tube;
$v^2 = 2ay$
$v = \sqrt{2ay}$
$v = \sqrt{(2)(30.2~m/s^2)(1.0~m)}$
$v = 7.77~m/s$
We can find the maximum height the ball reaches above the tube. We can let $v_0 = 7.77~m/s$ for this part of the question.
$y = \frac{v^2-v_0^2}{2g}$
$y = \frac{0-(7.77~m/s)^2}{(2)(-9.80~m/s^2)}$
$y = 3.08~m$
The ball reaches a height of 3.08 meters above the top of the tube.