Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 0.83 s after passing his window. Roger’s room is on the third floor, 15 m above the sidewalk. (a) How fast are the balloons traveling when they pass Roger’s window? (b) 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.

Acceleration is expressed as the quantity that indicates the relationship between the variations in the velocity of a particular object in unit time.

Given data.

Time, .$t=0.83\mathrm{s}$.

Distance$h=15\mathrm{m}$

The height of each floor is$d=5\mathrm{m}$

The relation from the equation of motion is given by.

$h=vt+\frac{1}{2}g{t}^2$

Here, g is the gravitational acceleration, v is the required speed, and t is the time.

On plugging the values in the above relation, you get.

$\begin{array}{c}15\mathrm{m}=v\left(0.83\mathrm{s}\right)+\frac{1}{2}\left(9.81\mathrm{m}/{\mathrm{s}}^2\right){\left(0.83\mathrm{s}\right)}^2\\ v=14.054\mathrm{m}/\mathrm{s}\end{array}$

Thus, $v=14.054\mathrm{m}/\mathrm{s}$is the required speed.

The relation to calculate the number of floors is given by.

$v^2=u^2+2gs$

Here, u is the initial velocity of the balloons whose value is zero, and s is the distance traveled by the balloon.

On plugging the values in the above relation, you get.

$\begin{array}{c}{\left(14.054\mathrm{m}/\mathrm{s}\right)}^2={\left(0\right)}^2+2\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)s\\ s=10.06\mathrm{m}\end{array}$

The number of floors is calculated as.

$\begin{array}{c}N=\frac{s}{d}\\ N=\left(\frac{10.06\mathrm{m}}{5\mathrm{m}}\right)\\ N\approx 2\end{array}$

The total number of floors can be calculated as.

$\begin{array}{l}{N}_t=2+2\\ N_t=4\end{array}$

Thus, $N_t=4$is the total number of floors.