Question

a group of students are conducting an experiment on free fall motion by dropping bricks from different heights out of the window of a tall building. during the first they drop a brick from a window that is 20 m high and it take t seconds to fall to the ground. the second trial they drop an identical brick from a window 80 m long will it take to fall to the ground? air resistance is negligible
a t seconds
b 2t seconds
c 4t seconds
d 16t seconds

Explanation:

Step1: Recall free - fall formula

For an object in free - fall (initial velocity \(u = 0\), acceleration \(a=g\)), the displacement \(h\) is given by the formula \(h=\frac{1}{2}gt^{2}\), where \(t\) is the time of fall.

For the first case, \(h_1 = 20\space m\) and \(t_1=T\space s\). So, \(20=\frac{1}{2}gT^{2}\) (Equation 1).

Step2: Analyze the second case

In the second case, \(h_2 = 80\space m\) and let the time of fall be \(t_2\). Then, \(80=\frac{1}{2}gt_2^{2}\) (Equation 2).

Step3: Divide Equation 2 by Equation 1

Divide the equation for the second case by the equation for the first case:
\[

$$\begin{align*} \frac{80}{20}&=\frac{\frac{1}{2}gt_2^{2}}{\frac{1}{2}gT^{2}}\\ 4&=\frac{t_2^{2}}{T^{2}} \end{align*}$$

\]

Step4: Solve for \(t_2\)

Take the square root of both sides of the equation \(4 = \frac{t_2^{2}}{T^{2}}\). Since time is positive, we have \(\frac{t_2}{T}=2\) (because \(\sqrt{4} = 2\)), so \(t_2 = 2T\).

Answer:

B. \(2T\) seconds