Answer
(a) Bruce should throw the bag with an initial speed of 12.9 m/s.
(b) Henrietta is 35.9 meters past the window when she catches the bagels.
Work Step by Step
We can find the time it takes the bag to fall 38.0 meters.
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{(2)(38.0~m)}{9.80~m/s^2}}$
$t = 2.785~s$
When the bag lands, Henrietta has been running for 9.00 seconds + 2.785 seconds which is 11.785 seconds. We can find Henrietta's distance $x$ from the window when the bagels land.
$x = v~t = (3.05~m/s)(11.785~s) = 35.9~m$
We can use this distance to find the initial speed $v_0$ of the bag.
$v_0 = \frac{x}{t} = \frac{35.9~m}{2.785~s} = 12.9~m/s$
(a) Bruce should throw the bag with an initial speed of 12.9 m/s.
(b) Henrietta is 35.9 meters past the window when she catches the bagels.
