Jittu Mathew - Physics
The ball released from a height (h) , will fall downwards ,only due to the gravitational force exerted by the Earth, if the air resistance is negligible. Therefore the acceleration acting on the ball will be acceleration due to gravity , which is , $a=9.8m/{s}^{2}$. The distance travelled by an object after a time period of $t=1s$ , can be calculated by using the formula ,$s\left(t\right)=ut+\frac{1}{2}a{t}^{2}$ , where $u=0m/s$ is the initial velocity as the object was at rest.
$\begin{array}{rcl}s\left(t\right)& =& ut+\frac{1}{2}a{t}^{2}\\ s\left(1\right)& =& 0\left(t\right)+\frac{1}{2}\times (9.8m/{s}^{2})\times {(1s)}^{2}\\ & =& \frac{9.8m}{2}\\ & =& 4.9m\end{array}$
This means that the vertical distance covered by the ball to reach the nineteenth floor window , or the vertical height of each floor, will be $s\left(1\right)=4.9m$.
The distance travelled by an object after a time period of $t=3s$ , can be calculated by using the formula ,$s\left(t\right)=ut+\frac{1}{2}a{t}^{2}$ , where $u=0m/s$ is the initial velocity as the object was at rest and the acceleration due to gravity ,$a=9.8m/{s}^{2}$.
$\begin{array}{rcl}s\left(t\right)& =& ut+\frac{1}{2}a{t}^{2}\\ s\left(3\right)& =& 0\left(t\right)+\frac{1}{2}\times (9.8m/{s}^{2})\times {(3s)}^{2}\\ & =& \frac{(9.8\times 9)m}{2}\\ & =& 44.1m\end{array}$
Thus the vertical distance covered by the ball in $t=3s$ will be $s\left(3\right)=44.1m$.
The vertical height of each floor, will be $s\left(1\right)=4.9m$ and the total distance travelled by the ball in $t=3s$ will be $s\left(3\right)=44.1m$.Thus for calculating the number of floors passed (n), the ratio of vertical distance travelled in $t=3s$ to the vertical distance travelled in $t=1s$ can be taken.
$\begin{array}{rcl}n& =& \frac{s\left(3\right)}{s\left(1\right)}\\ & =& \frac{44.1m}{4.9m}\\ & =& 9\end{array}$
Thus after $t=3s$ the ball will pass through 7 to 10 floors.
Answer: b) 7 to 10 floors.