Question

An object is dropped from rest out of the window of a building, and the time to hit the ground is found to be 5 seconds. The same object is then dropped from rest out of a window twice as high above the ground as the original window. The time it takes the object to hit the ground is closest to:

Answer 1

7 s

Explanation:

Given:

u = 0m/s

t = 5 sec

g = 9.8 m/s^2

The first thing is to calculate the height of the building, using the following formula:

`\begin{gathered} s=ut+(1)/(2)gt^2 \\ \text{ Replacing} \\ s=0\cdot5+(1)/(2)\cdot9.8\cdot5^2 \\ s=122.5\text{ m} \end{gathered}`

Now, we apply the same formula, but we substitute the double value of the distance and solve for t, just like this:

`\begin{gathered} 2\cdot122.5=(1)/(2)\cdot9.8\cdot\: t^2 \\ 9.8\cdot t^2=245\cdot2 \\ t^2=(490)/(9.8) \\ t=\sqrt[]{50} \\ t=7.07\text{ sec} \\ t\approx7\text{ sec} \end{gathered}`

The time it takes for the object to fall is 7 seconds.