Answer:
a) Speed of the balloons traveling when they pass Roger's window = 9.9 m/s
b) The balloons are being released from a floor away from Rogers' floor; so, the fourth floor.
Explanation:
We assume that the effects of air resistance are negligible.
Using the equations of motion,
g = 9.8 m/s², y = H = 15 m,
Initial velocity at his window level, u = ?
Time of fall (measured from his window to the ground), t = 1.01 s
a) y = ut + gt²/2
15 = u(1.01) + 9.8(1.01²)/2
1.01 u = 15 - 4.998
u = 10/1.01
u = 9.9 m/s
b) Starting from rest, initial velocity = u = 0 m/s, g = 9.8 m/s²
Now, let y = distance from where the balloons were released up to Rogers room.
v = 9.9 m/s
v² = u² + 2gy
9.9² = 0 + (2×9.8)y
y = 98.01/19.6 = 5 m
Each floor is 5 m high.
Since the distance from the balloons' release to Rogers' window is 5m, the balloons were released from the floor directly above his own, the fourth floor.