#### Answer

The minimum pushoff speed is 1.9 m/s. They are in the air for 2.7 seconds.

#### Work Step by Step

We can find an expression for the time it takes to fall a vertical distance $y$ using the formula:
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}$
We can then use the time t to find the minimum pushoff speed;
$v_x = \frac{x}{t}$
$v_x = x~\sqrt{\frac{g}{2y}}$
$v_x = (5.0~m)~\sqrt{\frac{9.80~m/s^2}{(2)(35~m)}}$
$v_x = 1.9~m/s$
We can find the time that they are in the air:
$t = \sqrt{\frac{2y}{g}}$
$t = \sqrt{\frac{(2)(35 ~)}{9.80 ~m/s^2}} = 2.7 ~s$