Answer
(a) The balloons are moving with a speed of 14 m/s when they pass Roger's window.
(b) The balloons are being released from the fifth floor.
Work Step by Step
(a) For this part of the problem, let $v_0$ be the speed of the balloons at Roger's window.
$y = v_0~t + \frac{1}{2}at^2$
$v_0 = \frac{y - \frac{1}{2}at^2}{t} = \frac{15 ~m - \frac{1}{2}(9.8 ~m/s^2)(0.83 ~s)^2}{0.83 ~s} = 14 ~m/s$
The balloons are moving with a speed of 14 m/s when they pass Roger's window.
(b) For this part of the problem, let v = 14 m/s.
$\Delta y = \frac{v^2 - v_0^2}{2a} = \frac{(14 ~m/s )^2}{(2)(9.8 ~m/s^2)} = 10 ~m$
The balloons are starting from a height 10 meters above Roger's window. The total height is 25 meters above the ground. Since each floor is 5.0 meters, the balloons are being released from the fifth floor.
