Question

Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 0.59 s after passing his window. Roger's room is on the second floor, 10 m above the sidewalk.

How fast are the balloons traveling when they pass Roger's window?

Assuming the balloons are being released from rest, from what floor are they being released? Each floor of the dorm is 5.0 m high.

Answer 1

Explanations & Calculations

• Assuming there is no air resistance acting on the balloon hence falling at g, you can apply $s = ut +\frac{1}{2}at^2$ vertically downward on the balloon.

$\qquad\qquad \begin{aligned} \small 10\,m&=\small v.(0.59\,s)+\frac{1}{2}(9.8\,ms^{-2})(0.59\,s)^2\\ \small v&=\small 14.1\,ms^{-1} \end{aligned}$

• For the last part, apply $\small v^2 =u^2+2as \downarrow$ from the released point to the second floor.

$\qquad\qquad \begin{aligned} \small (14.1\,ms^{-1})^2&=\small 0+2(9.8\,ms^{-2})(s)\\ \small s&=\small 10.14\,m\\ &\approx \small2(10\,m) \end{aligned}$

• Therefore, it has been released from 2 more floors above hence from the 4th floor.