# Question 4abae

Mar 12, 2017

$\text{4.24 s}$

#### Explanation:

The first thing you need to do here is to figure out the distance covered by the boy in $\text{3 s}$ of free falling by using the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{h = {v}_{0} \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}}}}$

Here

• $h$ is half of the height of the tower
• ${v}_{0}$ is the initial velocity of the boy, equal to ${\text{0 m s}}^{- 1}$ because the boy starts from rest
• $t$ is the time it takes for the boy to cover half of the height of the tower, equal to $\text{3 s}$
• $g$ is the gravitational acceleration, equal to ${\text{9.81 m s}}^{- 2}$

So, plug in your values to find

$h = \frac{1}{2} \cdot \text{9.81 m" color(red)(cancel(color(black)("s"^(-2)))) * 3^2 color(red)(cancel(color(black)("s"^2))) = "44.145 m}$

This means that the tower has a height of

$H = 2 \times h = 2 \times \text{44.145 m" = "88.29 m}$

Now all you have to do is use the same equation again, only this time looking to solve for ${t}_{\text{T}}$, the total time needed for the boy to fall a distance equal to $H$

$H = \frac{1}{2} \cdot g \cdot {t}_{\text{T"^2 implies t_"T}}^{2} = \sqrt{\frac{2 \cdot H}{g}}$

Plug in your values to find

t_"T" = sqrt( (2 * 88.29 color(red)(cancel(color(black)("m"))))/(9.81 color(red)(cancel(color(black)("m"))) "s"^(-2))) = color(darkgreen)(ul(color(black)("4.24 s")))#

I'll leave the answer rounded to three sig figs, but keep in mind that you only have one significant figure for the time it takes the ball to fall for half of the height of the tower.

Notice that it takes $\text{3 s}$ for the boy to fall for the first half of the tower and only

$\text{4.24 s " - " 3 s" = "1.24 s}$

for him to fall for the second height of the tower. This goes to show that his is being accelerated by the force of gravity, i.e. his velocity increases by ${\text{9.81 m s}}^{- 1}$ with every passing second of free fall.