Question

Figure 1.18 (Chapter 1) shows the Hoover Dam Bridge over

the Colorado River at a height of 271 m. If a heavy object is
dropped from the bridge, how much time passes before the
object makes a splash?

Answer 1

Answer: 7.436 s

Step-by-step explanation:

This situation is related to vertical motion, specifically free fall and can be modelled by the following equation:

`y=y_(o)+V_(o) t+(gt^(2))/(2)`

Where:

`y= 0m` is the final height of the object (when it makes splash)

`y_(o)=271 m` is the initial height of the object

`V_(o)=0 m/s` is the initial velocity of the object (it was dropped)

`g=-9.8m/s^(2)` is the acceleration due gravity (directed downwards)

`t` is the time since the objecct is dropped until it makes splash

`0=y_(o)+0+(gt^(2))/(2)`

Clearing
`t`:

`t=\sqrt{(-2y_(o))/(g)}`

`t=\sqrt{(-2(271 m))/(-9.8m/s^(2))}`

Finally:

`t=7.436 s`